## Projectile Motion

Projectile motionis motion under the influence ofgravity. If we stand at the edge of the roof of the Science Building and throw a ball up at an angle, it moves up and then down vertically while it moves horizontally.This is

projectile motion.To better understand this projectile motion, let's move back and then look at it through the eyes of two different and special observers.

What is the motion seen by a far-distant observer on the ground?

This observer is far enough away she has lost depth perception but can clearly see the ball rise and fall. She observes free fall, just as if the ball were thrown straight up. This is

vertical motion with constant acceleration.What motion is seen by an observer overhead? This overhead observer is high enough that he has lost depth perception but can clearly see the ball move horizontally. He observes

horizontal motion with constant velocity.Projectile motion, then, is a combination of vertical motion with constant acceleration (free fall that we have already discussed) and horizontal motion with constant velocity (which we also understand).

ExampleTo make the arithmetic easy, let's use the approximation that g = 10 m/s

^{2}and throw a ball from the top of the Science Building and look at its velocity and position.We throw the ball so it moves up with an initial vertical velocity of v

_{yo}= 20 m/s and so it moves horizontally with an initial horizontal velocity of v_{xo}= 15 m/s.We could also describe this as having an initial velocity of v

_{o}= 25 m/s at an angle of 53° from the horizontal. For angles measured from the horizontal, we knowv _{xo}= v_{o}cosv

_{yo}= v_{o}sinLook at the horizontal components; look at the v

_{x}'s. This is horizontal motion with constant velocity.Look at the vertical components; look at the v

_{y}'s. This is common, ordinary free fall; this is vertical velocity with constant acceleration.Now we will throw the ball yet another time. This time we will look at its position or its displacement.

These displacements come from the x- and y-component equations,

x = x _{o}+ v_{xo}t +^{1}/_{2}a_{x}t^{2}and

y = y _{o}+ v_{yo}t +^{1}/_{2}a_{y}t^{2}

Parabolic Trajectories

Water -- from a water fountain or a garden hose or a fire hose -- offers an example of

projectilemotionthat is easy to see.The

shapeof this path of water is aparabola.When a ball is in motion -- after being spiked or hit or thrown or kicked or dunked -- it undergoes

projectile motionand follows the path of aparabola.

Another Example

More ExamplesIt is fun and interesting to look at things like the

Maximum Heightor theMaximum Distance, called the HorizontalRange. But, please, treat these as interesting examples to be solved or derived rather than important formulas to be memorized.From a

level surface, howfarwill a projectile go?We begin by firing a projectile with initial velocity

v_{o}, that is with initial speed v_{o}at angle . It starts from the origin, (x_{o}= 0, y_{o}= 0). How far does it movehorizontallybefore it is back at its original vertical position (y = 0) again.While it may be more convenient or more common to describe the initial velocity

v_{o}in terms of its speed v_{o}and angle , it is easier -- and necessary -- to solve the equations in terms of the x- and y-components of the initial velocity.v _{xo}= v_{o}cosv

_{yo}= v_{o}sinTo find out

how farit goes, we must first find outhow longit is in the air. One way is to look at how long it takes to get to the top. At the top, the vertical component, the y-component, of the velocity is zero. We could find the time it takes for v_{y}to vanish. It will requiretwicethis time to come up and back down to the ground.Perhaps a more direct approach is just to find the time required to get back to the ground, to get back to y = 0. We know the vertical motion is explained by

y = y _{o}+ v_{yo}t +^{1}/_{2}a_{y}t^{2}where the acceleration is

a _{y}= - gand

v _{yo}= v_{o}sinand

y _{o}= 0Therefore

y = 0 + (v _{o}sin ) t -^{1}/_{2}g t^{2}We will set this equal to zero, y = 0, and solve for the time t,

(v _{o}sin ) t -^{1}/_{2}g t^{2}= 0[(v

_{o}sin ) t -^{1}/_{2}g t ] t = 0There are two solutions,

t = 0 and

t _{tot}= t = (2 v_{o}sin ) / gBoth these times are "true". The first, t = 0, refers to the initial time when the projectile

lefty = 0. The second, t_{tot}, refers to the time when the projectile getsback toy = 0; that is the one of interest.How far,

horizontally,has the projectile moved in this time?x = v _{x}tRange = X

_{max}= x(t_{tot}) = v_{xo}t_{tot}= (v_{o}cos ) (2 v_{o}sin ) / gX

_{max}= 2 v_{o}^{2}cos sin / gSince this expression contains

cos sin and

cos = sin (90 ^{o}- )this means that the range is the same for an angle and for its complimentary angle (90

^{o}-). That is, for the same initial speed v_{o}, initial angles of 30^{o}and 60^{o}provide the same range. Initial angles of 20^{o}and 70^{o}provide the same range.We can play a bit with trig identities and recall that

2 cos sin = sin 2 to rewrite the range equation as

X _{max}= v_{o}^{2}sin 2 / gFrom this form of the range equation, we can also see that the maximum value of the range occurs for

= 45 ^{o}since the maximum value of sin 2 is 1.0 when 2 is 90

^{o}, or = 45^{o}.

Another ExampleProjectile motion does not need to begin and end on a horizontal plane. We have already thrown a ball from the top of the Science Building and watched it hit the ground 25 m below. Now, let's throw a rock from the bottom of a cliff and ask where it lands on the ground above.

A rock is thrown with an initial speed of 30 m/s at an angle of 60

^{o}above the horizontal. As shown in the sketch, it is thrown from a canyon floor and goes up into the air and then lands on the plateau, 20 mabovethe canyon floor. Where does it hit the ground? Or, how far does it travel horizontally?Perhaps the very first thing to do is to restate the initial velocity in terms of its components; we know

v _{xo}= v_{o}cosv

_{xo}= (30 m/s) (0.50)v

_{xo}= 15 m/sand

v _{yo}= v_{o}sinv

_{yo}= (30 m/s) (0.866)v

_{yo}= 26 m/sBefore we can answer "how far" it travels, it is easier to ask "how long" is it in the air? We know its y-position is given by

y = y _{o}+ v_{yo}t +^{1}/_{2}a_{y}t^{2}We need the time for which the y-coordinate is at the plateau, y = 20 m.

20 m = 0 + (26 m/s) t + ^{1}/_{2}( - g ) t^{2}20 m = (26 m/s) t +

^{1}/_{2}( - 9.8 m/s^{2}) t^{2}20 m = (26 m/s) t - ( 4.9 m/s

^{2}) t^{2}20 = 26 t - 4.9 t

^{2}4.9 t

^{2}- 26 t + 20 = 0Solve this from the quadratic equation,

t _{1}= 0.9 st

_{2}= 4.4 sWhat is the meaning of these

twotimes? t_{1}is the time the rock passes the level of the plateau on its wayup. t_{2}is the time the rock is at the plateau on its waydown; t_{2}is when itlandson the plateau. That is the time we want. How far has it traveled horizontally during this time?X _{max}= v_{xo}t_{2}X

_{max}= (15 m/s) (4.4 s)

X_{max}= 66 m

2-D MotionCircular MotionReturn to ToC, Vectors and 2D Motion(c) 2002, Doug Davis; all rights reserved