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4 Move in Two and 3 Dimensions

4.3 Projectile Motion

Learning Objectives

By the end of this section, you will be able to:

Utilise one-dimensional motion in perpendicular directions to analyze projectile motion.
Summate the range, fourth dimension of flight, and maximum acme of a projectile that is launched and impacts a flat, horizontal surface.
Find the time of flight and impact velocity of a projectile that lands at a unlike height from that of launch.
Calculate the trajectory of a projectile.

Projectile motion is the motility of an object thrown or projected into the air, subject merely to acceleration as a result of gravity. The applications of projectile move in physics and engineering are numerous. Some examples include meteors as they enter Earth'south atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory. The movement of falling objects as discussed in Movement Along a Directly Line is a simple one-dimensional type of projectile motility in which in that location is no horizontal motion. In this department, we consider two-dimensional projectile motion, and our treatment neglects the furnishings of air resistance.

The most important fact to remember here is that motions forth perpendicular axes are independent and thus tin be analyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical and horizontal motions are contained. The key to analyzing two-dimensional projectile motion is to pause it into two motions: i along the horizontal centrality and the other forth the vertical. (This choice of axes is the most sensible considering acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we telephone call the horizontal axis the x-axis and the vertical axis the y-axis. Information technology is not required that we employ this choice of axes; it is simply user-friendly in the case of gravitational dispatch. In other cases we may choose a different fix of axes. (Figure) illustrates the note for displacement, where we ascertain

$\overset{\to }{s}$

to be the total displacement, and

$\overset{\to }{x}$

and

$\overset{\to }{y}$

are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are southward, x, and y.

An illustration of a soccer player kicking a ball. The soccer player's foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle theta between the x axis and s. — **Effigy iv.11** The total displacement s of a soccer ball at a point along its path. The vector
$\overset{\to }{s}$

has components

$\overset{\to }{x}$

and

$\overset{\to }{y}$

along the horizontal and vertical axes. Its magnitude is s and information technology makes an angle θ with the horizontal.

To draw projectile movement completely, we must include velocity and acceleration, too equally displacement. We must find their components along the x- and y-axes. Let's assume all forces except gravity (such every bit air resistance and friction, for example) are negligible. Defining the positive direction to be upwardly, the components of acceleration are then very unproblematic:

${a}_{y}=\text{−}g=-9.8\,\text{m}\text{/}{\text{s}}^{2}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}).$

Because gravity is vertical,

${a}_{x}=0.$

${a}_{x}=0,$

this ways the initial velocity in the x management is equal to the final velocity in the 10 direction, or

${v}_{x}={v}_{0x}.$

With these conditions on dispatch and velocity, we can write the kinematic (Equation) through (Equation) for motion in a uniform gravitational field, including the rest of the kinematic equations for a abiding acceleration from Motility with Abiding Acceleration. The kinematic equations for movement in a uniform gravitational field become kinematic equations with

${a}_{y}=\text{−}g,\enspace{a}_{x}=0:$

Horizontal Motility

${v}_{0x}={v}_{x},\,x={x}_{0}+{v}_{x}t$

Vertical Motion

$y={y}_{0}+\frac{1}{2}({v}_{0y}+{v}_{y})t$

${v}_{y}={v}_{0y}-gt$

$y={y}_{0}+{v}_{0y}t-\frac{1}{2}g{t}^{2}$

${v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0})$

Using this gear up of equations, we tin clarify projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

Resolve the motion into horizontal and vertical components along the ten– and y-axes. The magnitudes of the components of displacement
$\overset{\to }{s}$

along these axes are x and y. The magnitudes of the components of velocity

$\overset{\to }{v}$

are

${v}_{x}=v\text{cos}\,\theta \,\text{and}\,{v}_{y}=v\text{sin}\,\theta ,$

where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in (Figure).
Care for the motility equally 2 contained one-dimensional motions: 1 horizontal and the other vertical. Employ the kinematic equations for horizontal and vertical motion presented before.
Solve for the unknowns in the 2 separate motions: one horizontal and i vertical. Note that the only common variable between the motions is time t. The trouble-solving procedures here are the aforementioned as those for one-dimensional kinematics and are illustrated in the following solved examples.
Recombine quantities in the horizontal and vertical directions to find the full displacement
$\overset{\to }{s}$

and velocity

$\overset{\to }{v}.$

Solve for the magnitude and direction of the displacement and velocity using

$s=\sqrt{{x}^{2}+{y}^{2}},\enspace\theta ={\text{tan}}^{-1}(y\text{/}x),\enspace{v}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}},$

where θ is the direction of the displacement

$\overset{\to }{s}.$

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile's position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component. — **Figure 4.12** (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is unproblematic, because
${a}_{x}=0$

and

${v}_{x}$

is a constant. (c) The velocity in the vertical direction begins to decrease as the object rises. At its highest point, the vertical velocity is naught. As the object falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at whatsoever given point on the trajectory.

Example

A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of lxx.0 m/due south at an angle of

$75.0\text{°}$

above the horizontal, as illustrated in (Figure). The fuse is timed to ignite the vanquish just as it reaches its highest bespeak above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the vanquish and the explosion? (c) What is the horizontal displacement of the trounce when it explodes? (d) What is the full displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees. — **Figure 4.13** The trajectory of a fireworks shell. The fuse is fix to explode the shell at the highest point in its trajectory, which is found to be at a acme of 233 m and 125 m away horizontally.

Strategy

The movement can be cleaved into horizontal and vertical motions in which

${a}_{x}=0$

and

${a}_{y}=\text{−}g.$

We can then define

${x}_{0}$

and

${y}_{0}$

to be nil and solve for the desired quantities.

Solution

(a) By "summit" nosotros hateful the altitude or vertical position y in a higher place the starting point. The highest point in any trajectory, chosen the apex, is reached when

${v}_{y}=0.$

Since we know the initial and final velocities, equally well as the initial position, nosotros utilise the post-obit equation to find y:

${v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0}).$

Because

${y}_{0}$

and

${v}_{y}$

are both nothing, the equation simplifies to

$\text{0}={v}_{0y}^{2}-2gy.$

Solving for y gives

$y=\frac{{v}_{0y}^{2}}{2g}.$

Now we must find

${v}_{0y},$

the component of the initial velocity in the y direction. It is given by

${v}_{0y}={v}_{0}\text{sin}{\theta }_{0},$

where

${v}_{0}$

is the initial velocity of 70.0 m/s and

${\theta }_{0}=75\text{°}$

is the initial angle. Thus,

${v}_{0y}={v}_{0}\text{sin}\,\theta =(70.0\,\text{m}\text{/}\text{s})\text{sin}\,75\text{°}=67.6\,\text{m}\text{/}\text{s}$

and y is

$y=\frac{{(67.6\,\text{m}\text{/}\text{s})}^{2}}{2(9.80\,\text{m}\text{/}{\text{s}}^{2})}.$

Thus, we take

$y=233\,\text{m}\text{.}$

Annotation that because upwardly is positive, the initial vertical velocity is positive, every bit is the maximum height, just the acceleration resulting from gravity is negative. Note too that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.half-dozen-thou/s initial vertical component of velocity reaches a maximum height of 233 k (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which exercise reach such heights before exploding. In exercise, air resistance is non completely negligible, so the initial velocity would accept to exist somewhat larger than that given to reach the same height.

(b) Every bit in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this example, the easiest method is to use

${v}_{y}={v}_{0y}-gt.$

Because

${v}_{y}=0$

at the apex, this equation reduces to simply

$0={v}_{0y}-gt$

$t=\frac{{v}_{0y}}{g}=\frac{67.6\,\text{m}\text{/}\text{s}}{9.80\,\text{m}\text{/}{\text{s}}^{2}}=6.90\text{s}\text{.}$

This fourth dimension is also reasonable for big fireworks. If you are able to run into the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the fourth dimension is past using

$y\,\text{=}\,{y}_{0}+\frac{1}{2}({v}_{0y}+{v}_{y})t.$

This is left for you as an exercise to consummate.

${a}_{x}=0$

and the horizontal velocity is abiding, every bit discussed before. The horizontal displacement is the horizontal velocity multiplied by time equally given by

$x={x}_{0}+{v}_{x}t,$

where

${x}_{0}$

is equal to zero. Thus,

$x={v}_{x}t,$

where

${v}_{x}$

is the x-component of the velocity, which is given by

${v}_{x}={v}_{0}\text{cos}\,\theta =(70.0\,\text{m}\text{/}\text{s})\text{cos}75\text{°}=18.1\,\text{m}\text{/}\text{s}.$

Time t for both motions is the aforementioned, so x is

$x=(18.1\,\text{m}\text{/}\text{s})6.90\,\text{s}=125\,\text{m}\text{.}$

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found hither could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major event, and many fragments country directly below.

(d) The horizontal and vertical components of the deportation were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest betoken:

$\overset{\to }{s}=125\hat{i}+233\hat{j}$

$|\overset{\to }{s}|=\sqrt{{125}^{2}+{233}^{2}}=264\,\text{m}$

$\theta ={\text{tan}}^{-1}(\frac{233}{125})=61.8\text{°}.$

Note that the bending for the displacement vector is less than the initial bending of launch. To meet why this is, review (Figure), which shows the curvature of the trajectory toward the ground level.

When solving (Figure)(a), the expression nosotros found for y is valid for any projectile movement when air resistance is negligible. Telephone call the maximum pinnacle y = h. And then,

$h=\frac{{v}_{0y}^{2}}{2g}.$

This equation defines the maximum acme of a projectile above its launch position and information technology depends only on the vertical component of the initial velocity.

Cheque Your Understanding

A stone is thrown horizontally off a cliff

$100.0\,\text{m}$

high with a velocity of xv.0 m/southward. (a) Define the origin of the coordinate organization. (b) Which equation describes the horizontal move? (c) Which equations describe the vertical movement? (d) What is the rock's velocity at the point of impact?

[reveal-answer q="fs-id1165168031779″]Testify Solution[/reveal-answer]

[subconscious-answer a="fs-id1165168031779″]

(a) Choose the top of the cliff where the rock is thrown from the origin of the coordinate arrangement. Although it is arbitrary, nosotros typically choose time t = 0 to correspond to the origin. (b) The equation that describes the horizontal motion is

$x={x}_{0}+{v}_{x}t.$

With

${x}_{0}=0,$

this equation becomes

$x={v}_{x}t.$

${y}_{0}=0\,\text{and}\,{v}_{0y}=0,$

these equations simplify profoundly to become

$y=\frac{1}{2}({v}_{0y}+{v}_{y})t=\frac{1}{2}{v}_{y}t,\enspace$

${v}_{y}=\text{−}gt,\enspace$

$y=-\frac{1}{2}g{t}^{2},\enspace$

and

${v}_{y}^{2}=-2gy.$

(d) We employ the kinematic equations to detect the x and y components of the velocity at the betoken of bear on. Using

${v}_{y}^{2}=-2gy$

and noting the indicate of impact is −100.0 chiliad, we detect the y component of the velocity at impact is

${v}_{y}=44.3\,\text{m}\text{/}\text{s}.$

We are given the x component,

${v}_{x}=15.0\,\text{m}\text{/}\text{s},$

so we can calculate the full velocity at bear on: 5 = 46.8 m/southward and

$\theta =71.3\text{°}$

below the horizontal.
[/hidden-answer]

Example

Calculating Projectile Move: Tennis Player

A tennis histrion wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/south and at an bending

$45\text{°}$

above the horizontal ((Effigy)). On its way downward, the ball is caught past a spectator 10 m above the point where the brawl was striking. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball's velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball. — **Figure 4.14** The trajectory of a tennis ball striking into the stands.

Strategy

Again, resolving this 2-dimensional movement into ii independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motility lone. Thus, we solve for t kickoff. While the ball is ascent and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, nosotros recombine the vertical and horizontal results to obtain

$\overset{\to }{v}$

at terminal fourth dimension t, determined in the first part of the example.

Solution

(a) While the ball is in the air, it rises and then falls to a concluding position x.0 thou higher than its starting altitude. We can discover the time for this by using (Figure):

$y\,\text{=}\,{y}_{0}\,\text{+}\,{v}_{0y}t-\frac{1}{2}g{t}^{2}.$

If we take the initial position

${y}_{0}$

to exist zero, then the concluding position is y = 10 k. The initial vertical velocity is the vertical component of the initial velocity:

${v}_{0y}={v}_{0}\text{sin}\,{\theta }_{0}=(30.0\,\text{m}\text{/}\text{s})\text{sin}\,45\text{°}=21.2\,\text{m}\text{/}\text{s}.$

Substituting into (Figure) for y gives u.s.a.

$10.0\,\text{m}=(21.2\,\text{m/s})t-(4.90\,{\text{m/s}}^{\text{2}}){t}^{2}.$

Rearranging terms gives a quadratic equation in t:

$(4.90\,{\text{m/s}}^{\text{2}}){t}^{2}-(21.2\,\text{m/s})t+10.0\,\text{m}=0.$

Use of the quadratic formula yields t = iii.79 s and t = 0.54 s. Since the brawl is at a height of 10 one thousand at two times during its trajectory—in one case on the fashion up and once on the mode downwardly—nosotros take the longer solution for the time information technology takes the brawl to reach the spectator:

$t=3.79\,\text{s}\text{.}$

The time for projectile motion is adamant completely by the vertical motion. Thus, whatsoever projectile that has an initial vertical velocity of 21.2 g/south and lands 10.0 m below its starting altitude spends 3.79 s in the air.

(b) We tin can observe the concluding horizontal and vertical velocities

${v}_{x}$

and

${v}_{y}$

with the utilise of the result from (a). Then, we tin can combine them to find the magnitude of the total velocity vector

$\overset{\to }{v}$

and the angle

$\theta$

information technology makes with the horizontal. Since

${v}_{x}$

is abiding, we tin can solve for it at whatever horizontal location. Nosotros choose the starting point because we know both the initial velocity and the initial angle. Therefore,

${v}_{x}={v}_{0}\text{cos}{\theta }_{0}=(30\,\text{m}\text{/}\text{s})\text{cos}\,45\text{°}=21.2\,\text{m}\text{/}\text{s}.$

The final vertical velocity is given by (Figure):

${v}_{y}={v}_{0y}-gt.$

Since

${v}_{0y}$

was found in part (a) to exist 21.2 m/south, we have

${v}_{y}=21.2\,\text{m}\text{/}\text{s}-9.8\,\text{m}\text{/}{\text{s}}^{2}(3.79\,\text{s})=-15.9\,\text{m}\text{/}\text{s}.$

The magnitude of the final velocity

$\overset{\to }{v}$

$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{(21.2\,\text{m}\text{/}\text{s})}^{2}+{(\text{−}\,\text{15}.9\,\text{m}\text{/}\text{s})}^{2}}=26.5\,\text{m}\text{/}\text{s}.$

The management

${\theta }_{v}$

is found using the changed tangent:

${\theta }_{v}={\text{tan}}^{-1}(\frac{{v}_{y}}{{v}_{x}})={\text{tan}}^{-1}(\frac{21.2}{-15.9})=-53.1\text{°}.$

Significance

(a) As mentioned earlier, the fourth dimension for projectile motion is adamant completely by the vertical movement. Thus, any projectile that has an initial vertical velocity of 21.2 1000/s and lands x.0 m below its starting distance spends three.79 s in the air. (b) The negative angle means the velocity is

$53.1\text{°}$

below the horizontal at the point of bear upon. This effect is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity nosotros expect since information technology is impacting x.0 m above the launch height.

Time of Flight, Trajectory, and Range

Of involvement are the time of flying, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations requite useful expressions for these quantities, which are derived in the following sections.

Time of flight

Nosotros can solve for the time of flight of a projectile that is both launched and impacts on a apartment horizontal surface by performing some manipulations of the kinematic equations. We notation the position and deportation in y must be zero at launch and at impact on an fifty-fifty surface. Thus, nosotros set the displacement in y equal to zilch and find

$y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}g{t}^{2}=0.$

Factoring, we have

$t({v}_{0}\text{sin}{\theta }_{0}-\frac{gt}{2})=0.$

Solving for t gives us

${T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}{\theta }_{0})}{g}.$

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. (Figure) does not apply when the projectile lands at a unlike elevation than it was launched, as we saw in (Effigy) of the tennis thespian hitting the ball into the stands. The other solution, t = 0, corresponds to the fourth dimension at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to yard. Thus, on the Moon, where gravity is one-6th that of Globe, a projectile launched with the aforementioned velocity as on Earth would be airborne six times every bit long.

Trajectory

The trajectory of a projectile can be found by eliminating the fourth dimension variable t from the kinematic equations for arbitrary t and solving for y(x). We take

${x}_{0}={y}_{0}=0$

and then the projectile is launched from the origin. The kinematic equation for 10 gives

$x={v}_{0x}t⇒t=\frac{x}{{v}_{0x}}=\frac{x}{{v}_{0}\text{cos}{\theta }_{0}}.$

Substituting the expression for t into the equation for the position

$y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}g{t}^{2}$

gives

$y=({v}_{0}\text{sin}\,{\theta }_{0})(\frac{x}{{v}_{0}\text{cos}\,{\theta }_{0}})-\frac{1}{2}g{(\frac{x}{{v}_{0}\text{cos}\,{\theta }_{0}})}^{2}.$

Rearranging terms, we have

$y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}.$

This trajectory equation is of the form

$y=ax+b{x}^{2},$

which is an equation of a parabola with coefficients

$a=\text{tan}\,{\theta }_{0},\enspaceb=-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}.$

Range

From the trajectory equation we tin can also find the range, or the horizontal altitude traveled past the projectile. Factoring (Figure), we have

$y=x[\text{tan}\,{\theta }_{0}-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x].$

The position y is goose egg for both the launch bespeak and the impact bespeak, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions 10 = 0, corresponding to the launch betoken, and

$x=\frac{2{v}_{0}^{2}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{g},$

respective to the impact point. Using the trigonometric identity

$2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}2\theta$

and setting ten = R for range, we discover

$R=\frac{{v}_{0}^{2}\text{sin}2{\theta }_{0}}{g}.$

Annotation peculiarly that (Effigy) is valid simply for launch and bear on on a horizontal surface. We see the range is direct proportional to the square of the initial speed

${v}_{0}$

and

$\text{sin}2{\theta }_{0}$

, and it is inversely proportional to the dispatch of gravity. Thus, on the Moon, the range would be half dozen times greater than on Earth for the same initial velocity. Furthermore, we meet from the factor

$\text{sin}2{\theta }_{0}$

that the range is maximum at

$45\text{°}.$

These results are shown in (Figure). In (a) we meet that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at

$45\text{°}.$

This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to

$90\text{°}.$

The projectile launched with the smaller angle has a lower noon than the college angle, merely they both take the aforementioned range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle. — **Effigy 4.fifteen** Trajectories of projectiles on level ground. (a) The greater the initial speed
${v}_{0},$

the greater the range for a given initial angle. (b) The outcome of initial angle

${\theta }_{0}$

on the range of a projectile with a given initial speed. Notation that the range is the aforementioned for initial angles of

$15\text{°}$

and

$75\text{°},$

although the maximum heights of those paths are different.

Case

Comparing Golf Shots

A golfer finds himself in two dissimilar situations on different holes. On the second hole he is 120 m from the greenish and wants to hit the ball 90 g and let it run onto the green. He angles the shot depression to the ground at

$30\text{°}$

to the horizontal to allow the ball roll later on impact. On the fourth hole he is ninety m from the green and wants to allow the ball drop with a minimum corporeality of rolling subsequently affect. Here, he angles the shot at

$70\text{°}$

to the horizontal to minimize rolling afterward bear upon. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the ball at the 2nd hole?

(b) What is the initial speed of the ball at the fourth hole?

(d) Graph the trajectories.

Strategy

Nosotros see that the range equation has the initial speed and angle, so nosotros can solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

Solution

(a)

$R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}⇒{v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(9.8\,\text{m}\text{/}{\text{s}}^{2})}{\text{sin}(2(70\text{°}))}}=37.0\,\text{m}\text{/}\text{s}$

(b)

(c)

$\begin{array}{cc} y=x[\text{tan}\,{\theta }_{0}-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{Second hole:}\,y=x[\text{tan}\,70\text{°}-\frac{9.8\,\text{m}\text{/}{\text{s}}^{2}}{2{[(37.0\,\text{m}\text{/}\text{s)(}\text{cos}\,70\text{°})]}^{2}}x]=2.75x-0.0306{x}^{2}\hfill \\ \text{Fourth hole:}\,y=x[\text{tan}\,30\text{°}-\frac{9.8\,\text{m}\text{/}{\text{s}}^{2}}{2{[(31.9\,\text{m}\text{/}\text{s)(}\text{cos}30\text{°})]}^{2}}x]=0.58x-0.0064{x}^{2}\hfill \end{array}$

(d) Using a graphing utility, we can compare the ii trajectories, which are shown in (Figure).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees. — **Figure 4.16** Two trajectories of a golf ball with a range of ninety m. The impact points of both are at the same level as the launch point.

Significance

The initial speed for the shot at

$70\text{°}$

is greater than the initial speed of the shot at

$30\text{°}.$

Note from (Effigy) that 2 projectiles launched at the same speed only at different angles have the same range if the launch angles add together to

$90\text{°}.$

The launch angles in this example add to give a number greater than

$90\text{°}.$

Thus, the shot at

$70\text{°}$

has to have a greater launch speed to attain 90 yard, otherwise information technology would land at a shorter distance.

Check Your Understanding

If the two golf game shots in (Figure) were launched at the same speed, which shot would take the greatest range?

[reveal-answer q="fs-id1165166636799″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165166636799″]

The golf shot at

$30\text{°}.$

[/hidden-answer]

When we speak of the range of a projectile on level ground, nosotros presume R is very modest compared with the circumference of World. If, however, the range is big, Earth curves away below the projectile and the acceleration resulting from gravity changes management along the path. The range is larger than predicted by the range equation given earlier because the projectile has further to fall than it would on level ground, every bit shown in (Figure), which is based on a drawing in Newton's Principia . If the initial speed is not bad enough, the projectile goes into orbit. Earth'southward surface drops five m every 8000 m. In one s an object falls 5 one thousand without air resistance. Thus, if an object is given a horizontal velocity of

$8000\,\text{m}\text{/}\text{s}$

(or

$18,000\text{mi}\text{/}\text{hr})$

near Earth's surface, it volition go into orbit effectually the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when information technology was operational, or whatsoever satellite in a low World orbit. These and other aspects of orbital move, such as Globe'southward rotation, are covered in greater depth in Gravitation.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth. — **Figure 4.17** Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than information technology would be on level ground considering Earth curves away beneath its path. With a speed of 8000 m/s, orbit is achieved.

Summary

Projectile motion is the move of an object subject field only to the acceleration of gravity, where the acceleration is abiding, as near the surface of Earth.
To solve projectile move issues, we analyze the motion of the projectile in the horizontal and vertical directions using the ane-dimensional kinematic equations for ten and y.
The time of flight of a projectile launched with initial vertical velocity
${v}_{0y}$

on an fifty-fifty surface is given by

${T}_{tof}=\frac{2({v}_{0}\text{sin}\,\theta )}{g}.$

This equation is valid simply when the projectile lands at the same tiptop from which it was launched.
The maximum horizontal distance traveled past a projectile is chosen the range. Again, the equation for range is valid only when the projectile lands at the same top from which it was launched.

Conceptual Questions

Answer the following questions for projectile movement on level ground bold negligible air resistance, with the initial angle existence neither

$0\text{°}$

nor

$90\text{°}:$

(a) Is the velocity always nil? (b) When is the velocity a minimum? A maximum? (c) Tin can the velocity always be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the aforementioned as the initial speed at a time other than at t = 0?

[reveal-answer q="fs-id1165167780957″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165167780957″]

a. no; b. minimum at noon of trajectory and maximum at launch and bear upon; c. no, velocity is a vector; d. aye, where information technology lands

[/hidden-answer]

Answer the following questions for projectile movement on level ground bold negligible air resistance, with the initial angle beingness neither

$0\text{°}$

nor

$90\text{°}:$

(a) Is the acceleration ever null? (b) Is the dispatch e'er in the aforementioned management as a component of velocity? (c) Is the dispatch ever opposite in direction to a component of velocity?

A dime is placed at the border of a table so it hangs over slightly. A quarter is slid horizontally on the tabular array surface perpendicular to the border and hits the dime head on. Which coin hits the ground starting time?

[reveal-answer q="fs-id1165166623383″]Prove Solution[/reveal-reply]

[hidden-answer a="fs-id1165166623383″]

They both striking the basis at the same time.

[/hidden-answer]

Problems

A bullet is shot horizontally from shoulder height (1.5 thousand) with and initial speed 200 one thousand/s. (a) How much fourth dimension elapses earlier the bullet hits the footing? (b) How far does the bullet travel horizontally?

[reveal-reply q="fs-id1165168072758″]Testify Solution[/reveal-reply]

[hidden-answer a="fs-id1165168072758″]

$t=0.55\,\text{s}$

, b.

$x=110\,\text{m}$

[/hidden-answer]

A marble rolls off a tabletop 1.0 k high and hits the floor at a signal 3.0 thousand abroad from the table'due south edge in the horizontal direction. (a) How long is the marble in the air? (b) What is the speed of the marble when information technology leaves the table's edge? (c) What is its speed when it hits the flooring?

A sprint is thrown horizontally at a speed of 10 yard/s at the bull's-middle of a dartboard 2.4 m abroad, as in the following figure. (a) How far below the intended target does the dart hit? (b) What does your answer tell you lot near how proficient sprint players throw their darts?

[reveal-answer q="fs-id1165168078466″]Bear witness Solution[/reveal-reply]

[hidden-answer a="fs-id1165168078466″]

$t=0.24\text{s,}\enspaced=0.28\,\text{m}$

, b. They aim loftier.
An illustration of a person throwing a dart. The dart is released horizontally a distance of 2.4 meters from the dart board, level with the bulls eye of the dart board, with a speed of 10 meters per second.

[/hidden-answer]

An airplane flight horizontally with a speed of 500 km/h at a height of 800 thousand drops a crate of supplies (see the following figure). If the parachute fails to open, how far in front end of the release point does the crate hitting the basis?

An airplane releases a package. The airplane has a horizontal velocity of 500 kilometers per hour. The package's trajectory is the right half of a downward-opening parabola, initially horizontal at the airplane and curving down until it hits the ground.

Suppose the plane in the preceding problem fires a projectile horizontally in its direction of motion at a speed of 300 m/s relative to the plane. (a) How far in forepart of the release point does the projectile hitting the ground? (b) What is its speed when it hits the footing?

[reveal-answer q="fs-id1165167989106″]Bear witness Solution[/reveal-answer]

[hidden-answer a="fs-id1165167989106″]

a.,

$t=12.8\,\text{s,}\enspacex=5619\,\text{m}$

${v}_{y}=125.0\,\text{m}\text{/}\text{s,}\enspace{v}_{x}=439.0\,\text{m}\text{/}\text{s,}\enspace|\overset{\to }{v}|=456.0\,\text{m}\text{/}\text{s}$

[/hidden-answer]

A fastball pitcher tin throw a baseball at a speed of 40 m/s (90 mi/h). (a) Assuming the pitcher tin release the brawl 16.vii m from home plate then the brawl is moving horizontally, how long does it take the ball to achieve home plate? (b) How far does the brawl drib between the pitcher's paw and domicile plate?

A projectile is launched at an bending of

$30\text{°}$

and lands 20 s later on at the same height equally information technology was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Calculate the displacement from the bespeak of launch to the position on its trajectory at 15 s.

[reveal-answer q="fs-id1165166793284″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165166793284″]

${v}_{y}={v}_{0y}-gt,\enspacet=10\text{s,}\enspace{v}_{y}=0,\enspace{v}_{0y}=98.0\,\text{m}\text{/}\text{s},\enspace{v}_{0}=196.0\,\text{m}\text{/}\text{s}$

, b.

$h=490.0\,\text{m},$

${v}_{0x}=169.7\,\text{m}\text{/}\text{s,}\enspacex=3394.0\,\text{m,}$

$\begin{array}{cc} x=2545.5\,\text{m}\hfill \\ y=465.5\,\text{m}\hfill \\ \overset{\to }{s}=2545.5\,\text{m}\hat{i}+465.5\,\text{m}\hat{j}\hfill \end{array}$

[/hidden-answer]

A basketball player shoots toward a basket six.i m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an bending of

$60\text{°}$

above the horizontal, what must the initial speed be if it were to go through the handbasket?

At a detail instant, a hot air balloon is 100 m in the air and descending at a abiding speed of two.0 k/s. At this verbal instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 20 m/s. When she lands, where will she detect the ball? Ignore air resistance.

[reveal-reply q="fs-id1165166591102″]Show Solution[/reveal-respond]

[hidden-answer a="fs-id1165166591102″]

$-100\,\text{m}=(-2.0\,\text{m}\text{/}\text{s})t-(4.9\,\text{m}\text{/}{\text{s}}^{2}){t}^{2},$

$t=4.3\,\text{s,}$

$x=86.0\,\text{m}$

[/hidden-answer]

A human on a motorbike traveling at a compatible speed of x m/s throws an empty can directly upwards relative to himself with an initial speed of 3.0 m/southward. Notice the equation of the trajectory equally seen past a constabulary officeholder on the side of the road. Presume the initial position of the can is the point where it is thrown. Ignore air resistance.

An athlete tin can spring a distance of 8.0 m in the broad spring. What is the maximum distance the athlete can jump on the Moon, where the gravitational acceleration is 1-sixth that of Earth?

[reveal-answer q="fs-id1165167996165″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165167996165″]

${R}_{Moon}=48\,\text{m}$

[/hidden-answer]

The maximum horizontal distance a boy can throw a ball is 50 thousand. Presume he can throw with the same initial speed at all angles. How high does he throw the ball when he throws it straight upwards?

A rock is thrown off a cliff at an bending of

$53\text{°}$

with respect to the horizontal. The cliff is 100 m loftier. The initial speed of the rock is 30 1000/s. (a) How loftier in a higher place the edge of the cliff does the stone rise? (b) How far has it moved horizontally when it is at maximum altitude? (c) How long after the release does information technology hit the ground? (d) What is the range of the stone? (due east) What are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 s, t = 4.0 south, and t = vi.0 s?

[reveal-answer q="fs-id1165167746378″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165167746378″]

$\[\]$

${v}_{0y}=24\,\text{m}\text{/}\text{s}$

${v}_{y}^{2}={v}_{0y}^{2}-2gy⇒h=23.4\,\text{m}$

,
b.

$t=3\,\text{s}\enspace{v}_{0x}=18\,\text{m/s}\enspacex=54\,\text{m}$

$y=-100\,\text{m}\enspace{y}_{0}=0$

$y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}\enspace-100=24t-4.9{t}^{2}$

$⇒t=7.58\,\text{s}$

$x=136.44\,\text{m}$

$t=2.0\,\text{s}\enspacey=28.4\,\text{m}\enspacex=36\,\text{m}$

$t=4.0\,\text{s}\enspacey=17.6\,\text{m}\enspacex=22.4\,\text{m}$

$t=6.0\,\text{s}\enspacey=-32.4\,\text{m}\enspacex=108\,\text{m}$

[/hidden-reply]

Trying to escape his pursuers, a secret amanuensis skis off a slope inclined at

$30\text{°}$

beneath the horizontal at 60 km/h. To survive and land on the snowfall 100 m below, he must clear a gorge 60 m wide. Does he make it? Ignore air resistance.
A skier is moving with velocity v sub 0 down a slope that is inclined at 30 degrees to the horizontal. The skier is at the edge of a 60 m wide gap. The other side of the gap is 100 m lower.

A golfer on a fairway is 70 m away from the dark-green, which sits below the level of the fairway by xx g. If the golfer hits the brawl at an angle of

$40\text{°}$

with an initial speed of 20 m/s, how close to the green does she come?

[reveal-answer q="fs-id1165168065281″]Show Solution[/reveal-reply]

[hidden-respond a="fs-id1165168065281″]

${v}_{0y}=12.9\,\text{m}\text{/}\text{s}\,y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}\enspace-20.0=12.9t-4.9{t}^{2}$

$t=3.7\,\text{s}\enspace{v}_{0x}=15.3\,\text{m}\text{/}\text{s}⇒x=56.7\,\text{m}$

And so the golfer'south shot lands 13.3 one thousand short of the green.

[/hidden-respond]

A projectile is shot at a hill, the base of operations of which is 300 1000 away. The projectile is shot at

$60\text{°}$

above the horizontal with an initial speed of 75 m/s. The hill can be approximated by a aeroplane sloped at

$20\text{°}$

to the horizontal. Relative to the coordinate organisation shown in the following figure, the equation of this straight line is

$y=(\text{tan}20\text{°})x-109.$

Where on the loma does the projectile land?
A projectile is shot from the origin at a hill, the base of which is 300 m away. The projectile is shot at 60 degrees above the horizontal with an initial speed of 75 m/s. The hill is sloped away from the origin at 20 degrees to the horizontal. The slope is expressed as the equation y equals (tan of 20 degrees) times x minus 109.

An astronaut on Mars kicks a soccer brawl at an angle of

$45\text{°}$

with an initial velocity of 15 m/s. If the dispatch of gravity on Mars is 3.7 m/s, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same boot on the Moon, where gravity is one-sixth that of Earth?

[reveal-reply q="fs-id1165166572087″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165166572087″]

$R=60.8\,\text{m}$

,
b.

$R=137.8\,\text{m}$

[/hidden-answer]

Mike Powell holds the tape for the long bound of eight.95 k, established in 1991. If he left the ground at an angle of

$15\text{°},$

what was his initial speed?

MIT'south robot cheetah can leap over obstacles 46 cm high and has speed of 12.0 km/h. (a) If the robot launches itself at an angle of

$60\text{°}$

at this speed, what is its maximum tiptop? (b) What would the launch bending have to be to accomplish a superlative of 46 cm?

[reveal-answer q="fs-id1165167842253″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165167842253″]

${v}_{y}^{2}={v}_{0y}^{2}-2gy⇒y=2.9\,\text{m}\text{/}\text{s}$

$y=3.3\,\text{m}\text{/}\text{s}$

$y=\frac{{v}_{0y}^{2}}{2g}=\frac{{({v}_{0}\text{sin}\,\theta )}^{2}}{2g}⇒\text{sin}\,\theta =0.91⇒\theta =65.5\text{°}$

[/hidden-answer]

Mt. Asama, Japan, is an agile volcano. In 2009, an eruption threw solid volcanic rocks that landed ane km horizontally from the crater. If the volcanic rocks were launched at an angle of

$40\text{°}$

with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flying?

Drew Brees of the New Orleans Saints can throw a football 23.0 k/s (fifty mph). If he angles the throw at

$10\text{°}$

from the horizontal, what altitude does it go if information technology is to be caught at the aforementioned height as it was thrown?

[reveal-answer q="fs-id1165168098591″]Evidence Solution[/reveal-answer]

[subconscious-answer a="fs-id1165168098591″]

$R=18.5\,\text{m}$

[/hidden-answer]

The Lunar Roving Vehicle used in NASA's late Apollo missions reached an unofficial lunar land speed of 5.0 m/s by astronaut Eugene Cernan. If the rover was moving at this speed on a flat lunar surface and hitting a small bump that projected it off the surface at an angle of

$20\text{°},$

how long would it be "airborne" on the Moon?

A soccer goal is 2.44 thou high. A player kicks the ball at a altitude x m from the goal at an angle of

$25\text{°}.$

What is the initial speed of the soccer brawl?

[reveal-answer q="fs-id1165167854326″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1165167854326″]

$y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}⇒{v}_{0}=16.4\,\text{m}\text{/}\text{s}$

[/hidden-reply]

Olympus Mons on Mars is the largest volcano in the solar system, at a meridian of 25 km and with a radius of 312 km. If you are continuing on the meridian, with what initial velocity would you have to burn down a projectile from a cannon horizontally to articulate the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of

$3.7\,\text{m}\text{/}{\text{s}}^{2}.$

In 1999, Robbie Knievel was the offset to spring the Grand Coulee on a motorcycle. At a narrow part of the canyon (69.0 m wide) and traveling 35.8 m/southward off the takeoff ramp, he reached the other side. What was his launch angle?

[reveal-answer q="fs-id1165168009639″]Show Solution[/reveal-answer]

[subconscious-respond a="fs-id1165168009639″]

$R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}⇒{\theta }_{0}=15.0\text{°}$

[/hidden-reply]

Yous throw a baseball at an initial speed of 15.0 m/s at an angle of

$30\text{°}$

with respect to the horizontal. What would the ball'southward initial speed have to be at

$30\text{°}$

on a planet that has twice the dispatch of gravity equally Globe to achieve the same range? Consider launch and bear upon on a horizontal surface.

Aaron Rogers throws a football at 20.0 m/s to his broad receiver, who runs straight downwardly the field at 9.4 m/s for 20.0 m. If Aaron throws the football game when the broad receiver has reached 10.0 one thousand, what angle does Aaron have to launch the ball so the receiver catches information technology at the 20.0 grand mark?

[reveal-answer q="fs-id1165166777489″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165166777489″]

It takes the broad receiver 1.1 s to comprehend the terminal 10 grand of his run.

${T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}\,\theta )}{g}⇒\text{sin}\,\theta =0.27⇒\theta =15.6\text{°}$

[/hidden-respond]

Glossary

projectile motion: movement of an object subject merely to the acceleration of gravity

range: maximum horizontal distance a projectile travels

time of flying: elapsed time a projectile is in the air

trajectory: path of a projectile through the air

ivesholleatunce.blogspot.com

Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/4-3-projectile-motion/

Incorrect; Try Again; 3 Attempts Remaining Enter Your Answer Using Dimensions of Plane Angle

4.3 Projectile Motion

Learning Objectives

Problem-Solving Strategy: Projectile Motion

Example

A Fireworks Projectile Explodes High and Away

Strategy

Solution

Cheque Your Understanding

Example

Calculating Projectile Move: Tennis Player

Strategy

Solution

Significance

Time of Flight, Trajectory, and Range

Time of flight

Trajectory

Range

Case

Comparing Golf Shots

Strategy

Solution

Significance

Check Your Understanding

Summary

Conceptual Questions

Problems

Glossary

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