College Physics for AP® Courses

# 3.4Projectile Motion

### Learning Objectives

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

• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
• Determine the location and velocity of a projectile at different points in its trajectory.
• Apply the principle of independence of motion to solve projectile motion problems.

The information presented in this section supports the following AP® learning objectives:

• 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
• 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.36 illustrates the notation for displacement, where $ss size 12{s} {}$ is defined to be the total displacement and $xx size 12{x} {}$ and $yy size 12{y} {}$ are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation $AA size 12{A} {}$ to represent a vector with components $AxAx size 12{A rSub { size 8{x} } } {}$ and $AyAy size 12{A rSub { size 8{y} } } {}$. If we continued this format, we would call displacement $ss size 12{s} {}$ with components $sxsx size 12{s rSub { size 8{x} } } {}$ and $sysy size 12{s rSub { size 8{y} } } {}$. However, to simplify the notation, we will simply represent the component vectors as $xx size 12{x} {}$ and $yy size 12{y} {}$.)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: $ay=–g=–9.80 m/s2ay=–g=–9.80 m/s2 size 12{a rSub { size 8{y} } ="-g"="-9.80" "m/s" rSup { size 8{2} } } {}$. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, $ax=0ax=0 size 12{a rSub { size 8{x} } } {}$. Both accelerations are constant, so the kinematic equations can be used.

### Review of Kinematic Equations (constant $aa$)

$x = x 0 + v - t x = x 0 + v - t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}$
3.28
$v - = v 0 + v 2 v - = v 0 + v 2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {}$
3.29
$v = v 0 + at v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}$
3.30
$x = x 0 + v 0 t + 1 2 at 2 x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}$
3.31
$v 2 = v 0 2 + 2a ( x − x 0 ) . v 2 = v 0 2 + 2a ( x − x 0 ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a $$x - x rSub { size 8{0} }$$ } {}$
3.32
Figure 3.36 The total displacement $ss size 12{s} {}$ of a soccer ball at a point along its path. The vector $ss size 12{s} {}$ has components $xx size 12{x} {}$ and $yy size 12{y} {}$ along the horizontal and vertical axes. Its magnitude is $ss size 12{s} {}$, and it makes an angle $θθ size 12{θ} {}$ with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so $Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}$ and $Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {}$ are used. The magnitude of the components of displacement $ss size 12{s} {}$ along these axes are $xx size 12{x} {}$ and $y.y. size 12{y} {}$ The magnitudes of the components of the velocity $vv size 12{v} {}$ are $vx=vcosθvx=vcosθ size 12{v rSub { size 8{x} } =v"cos"θ} {}$ and $vy=vsinθ,vy=vsinθ, size 12{v rSub { size 8{y} } =v"sin"θ} {}$ where $vv size 12{v} {}$ is the magnitude of the velocity and $θθ size 12{θ} {}$ is its direction, as shown in Figure 3.37. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

$Horizontal Motion ( a x = 0 ) Horizontal Motion ( a x = 0 ) size 12{"Horizontal Motion " $$a rSub { size 8{x} } =0$$ } {}$
3.33
$x = x 0 + v x t x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}$
3.34
$vx=v0x=vx=velocity is a constant.vx=v0x=vx=velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}$
3.35
$Vertical Motion ( assuming positive is up a y = − g = − 9. 80 m/s 2 ) Vertical Motion ( assuming positive is up a y = − g = − 9. 80 m/s 2 ) size 12{"Vertical Motion " $$"assuming positive is up "a rSub { size 8{y} } = - g= - 9/"80"" m/s" rSup { size 8{2} }$$ } {}$
3.36
$y = y 0 + 1 2 ( v 0y + v y ) t y = y 0 + 1 2 ( v 0y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } $$v rSub { size 8{0y} } +v rSub { size 8{y} }$$ t} {}$
3.37
$v y = v 0y − gt v y = v 0y − gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}$
3.38
$y = y 0 + v 0y t − 1 2 gt 2 y = y 0 + v 0y t − 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}$
3.39
$vy2=v 0y 2−2g(y−y0).vy2=v 0y 2−2g(y−y0). size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g $$y - y rSub { size 8{0} }$$ "."} {}$
3.40

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time $tt size 12{t} {}$. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement $ss size 12{s} {}$ and velocity $vv size 12{v} {}$. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing $A = A x 2 + A y 2 A = A x 2 + A y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}$ and $θ=tan−1(Ay/Ax)θ=tan−1(Ay/Ax) size 12{θ="tan" rSup { size 8{ - 1} } $$A rSub { size 8{y} } /A rSub { size 8{x} }$$ } {}$ in the following form, where $θθ size 12{θ} {}$ is the direction of the displacement $ss size 12{s} {}$ and $θvθv size 12{θ rSub { size 8{v} } } {}$ is the direction of the velocity $vv size 12{v} {}$:

Total displacement and velocity

$s = x 2 + y 2 s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}$
3.41
$θ = tan − 1 ( y / x ) θ = tan − 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } $$y/x$$ } {}$
3.42
$v = v x 2 + v y 2 v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}$
3.43
$θv=tan−1(vy/vx).θv=tan−1(vy/vx). size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } $$v rSub { size 8{y} } /v rSub { size 8{x} }$$ "."} {}$
3.44
Figure 3.37 (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 simple, because $ax=0ax=0 size 12{a rSub { size 8{x} } =0} {}$ and $vxvx size 12{v rSub { size 8{x} } } {}$ is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the 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 any given point on the trajectory.

### Example 3.4A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of $75.0°75.0°$ above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Strategy

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which $ax=0ax=0 size 12{ a rSub { size 8{x} } =0} {}$ and $ay=–gay=–g size 12{ a rSub { size 8{y} } =-g} {}$. We can then define $x0x0 size 12{x rSub { size 8{0} } } {}$ and $y0y0 size 12{y rSub { size 8{0} } } {}$ to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position $yy size 12{y} {}$ above the starting point. The highest point in any trajectory, called the apex, is reached when $vy=0vy=0 size 12{ v rSub { size 8{y} } =0} {}$. Since we know the initial and final velocities as well as the initial position, we use the following equation to find $yy size 12{y} {}$:

$vy2=v0y2−2g(y−y0).vy2=v0y2−2g(y−y0). size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g $$y - y rSub { size 8{0} }$$ "."} {}$
3.45
Figure 3.38 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because $y0y0 size 12{y rSub { size 8{0} } } {}$ and $vyvy size 12{v rSub { size 8{y} } } {}$ are both zero, the equation simplifies to

$0=v 0y 2−2gy.0=v 0y 2−2gy. size 12{0=v rSub { size 8{0y} } rSup { size 8{2} } - 2 ital "gy."} {}$
3.46

Solving for $yy size 12{y} {}$ gives

$y = v 0y 2 2g . y = v 0y 2 2g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}$
3.47

Now we must find $v0yv0y size 12{v rSub { size 8{0y} } } {}$, the component of the initial velocity in the y-direction. It is given by $v0y=v0sinθv0y=v0sinθ size 12{v rSub { size 8{0y rSup} =v rSub {0 rSup size 12{"sin"θ}} {}$, where $v0yv0y$ is the initial velocity of 70.0 m/s, and $θ0=75.0°θ0=75.0° size 12{θ rSub { size 8{0} } } {}$ is the initial angle. Thus,

$v0y=v0sinθ0=(70.0 m/s)(sin 75°)=67.6 m/s.v0y=v0sinθ0=(70.0 m/s)(sin 75°)=67.6 m/s. size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } = $$"70" "." 0" m/s"$$ $$"sin""75" { size 12{ circ } }$$ ="67" "." 6" m/s."} {}$
3.48

and $yy size 12{y} {}$ is

$y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , size 12{y= { { $$"67" "." 6" m/s"$$ rSup { size 8{2} } } over {2 $$9 "." "80"" m/s" rSup { size 8{2} }$$ } } } {}$
3.49

so that

$y=233 m.y=233 m. size 12{y="233"" m."} {}$
3.50

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use $y=y0+12(v0y+vy)ty=y0+12(v0y+vy)t size 12{y=y rSub { size 8{0} } + { {1} over {2} } $$v rSub { size 8{0y} } +v rSub { size 8{y} }$$ t} {}$. Because $y0y0 size 12{y rSub { size 8{0} } } {}$ is zero, this equation reduces to simply

$y = 1 2 ( v 0y + v y ) t . y = 1 2 ( v 0y + v y ) t . size 12{y= { {1} over {2} } $$v rSub { size 8{0y} } +v rSub { size 8{y} }$$ t} {}$
3.51

Note that the final vertical velocity, $vyvy size 12{v rSub { size 8{y} } } {}$, at the highest point is zero. Thus,

t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s. t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s. alignl { stack { size 12{t= { {2y} over { $$v rSub { size 8{0y} } +v rSub { size 8{y} }$$ } } = { {2 times "233"" m"} over { $$"67" "." 6" m/s"$$ } } } {} # =6 "." "90"" s" {} } } {}
3.52

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using $y=y0+v0yt−12gt2y=y0+v0yt−12gt2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}$, and solving the quadratic equation for $tt size 12{t} {}$.)

Solution for (c)

Because air resistance is negligible, $ax=0ax=0 size 12{a rSub { size 8{x} } =0} {}$ and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by $x=x0+vxtx=x0+vxt size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}$, where $x0x0 size 12{x rSub { size 8{0} } } {}$ is equal to zero:

$x=vxt,x=vxt, size 12{x=v rSub { size 8{x} } t ","} {}$
3.53

where $vxvx size 12{v rSub { size 8{x} } } {}$ is the x-component of the velocity, which is given by $vx=v0cosθ0.vx=v0cosθ0. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } "." } {}$ Now,

$vx=v0cosθ0=(70.0 m/s)(cos 75.0°)=18.1 m/s.vx=v0cosθ0=(70.0 m/s)(cos 75.0°)=18.1 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 12{0} } = $$"70" "." 0" m/s"$$ $$"cos""75.0°"$$ ="18" "." 1" m/s."} {}$
3.54

The time $tt size 12{t} {}$ for both motions is the same, and so $xx size 12{t} {}$ is

$x=(18.1 m/s)(6.90 s)=125 m.x=(18.1 m/s)(6.90 s)=125 m. size 12{x= $$"18" "." 1" m/s"$$ $$6 "." "90"" s"$$ ="125"" m."} {}$
3.55

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for $yy size 12{y} {}$ is valid for any projectile motion where air resistance is negligible. Call the maximum height $y=hy=h size 12{y=h} {}$; then,

$h = v 0 y 2 2 g . h = v 0 y 2 2 g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}$
3.56

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

### Defining a Coordinate System

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the $xx size 12{x} {}$ and $yy size 12{y} {}$ positions. Often, it is convenient to choose the initial position of the object as the origin such that $x0=0x0=0 size 12{x rSub { size 8{0} } =0} {}$ and $y0=0y0=0 size 12{y rSub { size 8{0} } =0} {}$. It is also important to define the positive and negative directions in the $xx size 12{x} {}$ and $yy size 12{y} {}$ directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration, $gg size 12{g} {}$, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, $gg size 12{g} {}$ takes a positive value.

### Example 3.5Calculating Projectile Motion: Hot Rock Projectile

Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle $35.0°35.0° size 12{"35"°} {}$ above the horizontal, as shown in Figure 3.39. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock's velocity at impact?

Figure 3.39 The trajectory of a rock ejected from the Kilauea volcano.

Strategy

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for $tt size 12{t} {}$ first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain $vv size 12{v} {}$ and $θvθv size 12{θ rSub { size 8{v} } } {}$ at the final time $tt size 12{t} {}$ determined in the first part of the example.

Solution for (a)

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

$y=y0+v0yt−12gt2 .y=y0+v0yt−12gt2 . size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } "."} {}$
3.57

If we take the initial position $y0y0 size 12{y rSub { size 8{0} } } {}$ to be zero, then the final position is $y=−20.0 m.y=−20.0 m. size 12{y= - "20" "." 0" m" "." } {}$ Now the initial vertical velocity is the vertical component of the initial velocity, found from $v0y=v0sinθ0v0y=v0sinθ0 size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } } {}$ = ()($sin 35.0°sin 35.0° size 12{"sin 35"°} {}$) = . Substituting known values yields

$−20.0 m=(14.3 m/s)t−4.90 m/s2t2 .−20.0 m=(14.3 m/s)t−4.90 m/s2t2 . size 12{ - "20" "." 0" m"= $$"14" "." 3" m/s"$$ t - left (4 "." "90"" m/s" rSup { size 8{2} } right )t rSup { size 8{2} } "."} {}$
3.58

Rearranging terms gives a quadratic equation in $tt size 12{t} {}$:

$4.90 m/s2t2−14.3 m/st−20.0 m=0.4.90 m/s2t2−14.3 m/st−20.0 m=0. size 12{ left (4 "." "90"" m/s" rSup { size 8{2} } right )t rSup { size 8{2} } - left ("14" "." "3 m/s" right )t - left ("20" "." 0" m" right )=0.} {}$
3.59

This expression is a quadratic equation of the form $at2 + bt + c = 0 at2 + bt + c = 0 size 12{ ital "at" rSup { size 8{2} } + ital "bt"+c=0} {}$, where the constants are $a=4.90 a=4.90$, $b=–14.3 b=–14.3$, and $c=–20.0. c=–20.0.$ Its solutions are given by the quadratic formula:

$t = − b ± b 2 − 4 ac 2a . t = − b ± b 2 − 4 ac 2a . size 12{t= { { - b +- sqrt {b rSup { size 8{2} } - 4 ital "ac"} } over {"2a"} } "." } {}$
3.60

This equation yields two solutions: $t=3.96t=3.96 size 12{t=3 "." "96"} {}$ and $t=–1.03t=–1.03 size 12{t=3 "." "96"} {}$. (It is left as an exercise for the reader to verify these solutions.) The time is $t=3.96st=3.96s size 12{t=3 "." "96""s"} {}$ or $–1.03s–1.03s size 12{-1 "." "03""s"} {}$. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

$t=3.96 s.t=3.96 s. size 12{t=3 "." "96"" s."} {}$
3.61

Discussion for (a)

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Solution for (b)

From the information now in hand, we can find the final horizontal and vertical velocities $vxvx size 12{v rSub { size 8{x} } } {}$ and $vyvy size 12{v rSub { size 8{y} } } {}$ and combine them to find the total velocity $vv size 12{v} {}$ and the angle $θ0θ0 size 12{θ rSub { size 8{0} } } {}$ it makes with the horizontal. Of course, $vxvx size 12{v rSub { size 8{x} } } {}$ is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

$vx=v0cosθ0=(25.0 m/s)(cos 35°)=20.5 m/s.vx=v0cosθ0=(25.0 m/s)(cos 35°)=20.5 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } = $$"25" "." 0" m/s"$$ $$"cos""35" rSup { size 8{ circ } }$$ ="20" "." 5" m/s."} {}$
3.62

The final vertical velocity is given by the following equation:

$vy=v0y−gt,vy=v0y−gt, size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt,"} {}$
3.63

where $v0yv0y size 12{v rSub { size 8{0y} } } {}$ was found in part (a) to be . Thus,

$v y = 14 . 3 m/s − ( 9 . 80 m/s 2 ) ( 3 . 96 s ) v y = 14 . 3 m/s − ( 9 . 80 m/s 2 ) ( 3 . 96 s ) size 12{v rSub { size 8{y} } ="14" "." 3" m/s" - $$9 "." "80"" m/s" rSup { size 8{2} }$$ $$3 "." "96"" s"$$ } {}$
3.64

so that

$vy=−24.5 m/s.vy=−24.5 m/s. size 12{v rSub { size 8{y} } = - "24" "." 5" m/s."} {}$
3.65

To find the magnitude of the final velocity $vv size 12{v} {}$ we combine its perpendicular components, using the following equation:

$v=vx2+vy2=(20.5 m/s)2+(−24.5 m/s)2,v=vx2+vy2=(20.5 m/s)2+(−24.5 m/s)2, size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } = sqrt { $$"20" "." 5" m/s"$$ rSup { size 8{2} } + $$- "24" "." 5" m/s"$$ rSup { size 8{2} } } ","} {}$
3.66

which gives

$v=31.9 m/s.v=31.9 m/s. size 12{v="31" "." 9" m/s."} {}$
3.67

The direction $θvθv size 12{θ rSub { size 8{v} } } {}$ is found from the equation:

$θ v = tan − 1 ( v y / v x ) θ v = tan − 1 ( v y / v x ) size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } $$v rSub { size 8{y} } /v rSub { size 8{x} }$$ } {}$
3.68

so that

$θv=tan−1(−24.5/20.5)=tan−1(−1.19).θv=tan−1(−24.5/20.5)=tan−1(−1.19). size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } $$- "24" "." 5/"20" "." 5$$ ="tan" rSup { size 8{ - 1} } $$- 1 "." "19"$$ "."} {}$
3.69

Thus,

$θv=−50.1°.θv=−50.1°. size 12{θ rSub { size 8{v} } = - "50" "." 1 rSup { size 12{ circ } "."} } {}$
3.70

Discussion for (b)

The negative angle means that the velocity is $50.1°50.1° size 12{"50" "." 1°} {}$ below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.39.)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance $RR size 12{R} {}$ traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

Figure 3.40 Trajectories of projectiles on level ground. (a) The greater the initial speed $v0v0 size 12{v rSub { size 8{0} } } {}$, the greater the range for a given initial angle. (b) The effect of initial angle $θ0θ0 size 12{θ rSub { size 8{0} } } {}$ on the range of a projectile with a given initial speed. Note that the range is the same for $15°15° size 12{"15"°} {}$ and $75°75° size 12{"75°"} {}$, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed $v0v0 size 12{v rSub { size 8{0} } } {}$, the greater the range, as shown in Figure 3.40(a). The initial angle $θ0θ0 size 12{θ rSub { size 8{0} } } {}$ also has a dramatic effect on the range, as illustrated in Figure 3.40(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately $38°38° size 12{"38°"} {}$. Interestingly, for every initial angle except $45°45° size 12{"45°"} {}$, there are two angles that give the same range—the sum of those angles is $90°90° size 12{"90°"} {}$. The range also depends on the value of the acceleration of gravity $gg size 12{g} {}$. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range $RR size 12{R} {}$ of a projectile on level ground for which air resistance is negligible is given by

$R=v02sin2θ0g,R=v02sin2θ0g, size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } ","} {}$
3.71

where $v0v0 size 12{v rSub { size 8{0} } } {}$ is the initial speed and $θ0θ0 size 12{θ rSub { size 8{0} } } {}$ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that $RR size 12{R} {}$ is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.41.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

Figure 3.41 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 it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

### PhET Explorations: Projectile Motion

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

Figure 3.42 Projectile Motion