Projectile Motion : Practice Problems + Solutions

Solve the following questions using what you know about projectile motion.

1.      A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s.

a)      What are the components of this velocity?

V_x = 3.5 m/s                            V_y = 0 m/s

b)      What will be the horizontal velocity 2 seconds after the bird leaves the cliff?

3.5 m/s – horizontal velocity is unchanging

c)      If the cliff is 300 m high, at what time will the roadrunner reach the ground?

h = d_y = ½ * 10 * t² = 300

300 * 2 / 10 = t² = 60

t = 7.75 s

d)     How far from the cliff will this bird land?

d_x = 3.5 * 7.75 = 27.125 m

e)      If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the roadrunner land in the pond?

Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.

f)       What is the final vertical velocity at which the roadrunner is traveling? [The vertical velocity at the time when the bird reaches the ground]

V_y = 10 * 7.75 + 0 = 77.5 m/s

g)      What is the final horizontal velocity at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]

V_x = 0 + 3.5 = 3.5 m/s

h)      What is the total final velocity of this motion? [magnitude and direction]

V² = 77.5² + 3.5² = 6018.5

V = 77.579 m/s

q = tan^-1 (77.5 / 3.5) = 87.41^o below the horizontal

2.      An object (any object) is dropped from a height of 300m

a)      How long does it take this object to fall to the ground?

h = d_y = ½ * 10 * t² = 300

300 * 2 / 10 = t² = 60

t = 7.75 s

b)      Compare this answer with you answer from question 1, part c). What are the reasons for any similarities or differences?

They are the same. This is because their vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.

3.      The intent of a bean bag toss game is to get your bean bag to land on the ‘bull’s-eye’ of a target. The target is set up parallel to the ground and is the same height above the ground as your hand is when you let go of the bean bag. The game’s rules further require you to be 5 m from the center of the target when you release the bag.

a)      Evaluate the following questions for both an angle of 32^o and an angle of 58^o if the bean bag is thrown with an initial velocity of 6 m/s:

                  32^o                                                                   58^o

i.        What are the components of velocity?

            V_x32: Cos 32^o = V_x32/6                                                V_x58: Cos 58^o = V_x58/6           

            6 * Cos 32^o = V_x32 = 5.09 m/s                                    6 * Cos 58^o = V_x58 = 3.18 m/s

            V_y32: Sin 32^o = V_y32/6                                                 V_y58: Sin 58^o = V_y58/6

            6 * Sin 32^o = V_y32 = 3.18 m/s                                     6 * Sin 58^o = V_y58 = 5.09 m/s

ii.      What is the maximum height of the bean bag’s motion?

            t_TOP32 = V_y/10 = 3.18/10 = 0.318s                   t_TOP58 = V_y58/10 = 5.09/10 = 0.509s

            h_MAX32 = ½ * 10 * 0.318² = 0.506 m               h_MAX58 = ½ * 10 * 0.509² = 1.295 m

iii.    How long will the bean bag be in the air?

            t_TOTAL32 = 2 * t_TOP32 = 0.636 s                        t_TOTAL58 = 2 * t_TOP58 = 1.018 s

iv.    How far away from you will the bag land?

            d_x32 = 5.09 * 0.636 =   3.24 m                         d_x58 = 3.18 * 1.018 = 3.24m

v.      If the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does your bean bag win?

                        No                                                                   No

4.      A stunt driver drives a red mustang convertible up a ramp and off a cliff. The car leaves the ramp at a velocity of 60 m/s at an angle of 45^o to the horizontal; the cliff and ramp combined cause the car to begin its projectile motion at a height of 315m above the ground. If you were coordinating this stunt, how far away would you put a landing surface so that your stunt driver was not injured?

First let’s think about strategy. The question is basically asking how far away from the cliff the car will land. In order to find horizontal distance we need horizontal velocity and time. We can find both horizontal and vertical velocity from the initial conditions, but we’ll have to calculate the time it will take for the car to reach the ground. So first we’ll find the components of velocity:

            V_x: Cos 45^o = V_x/60                                       V_y: Sin 45^o = V_y/60

            60 * Cos 45^o = V_x = 42.43 m/s                       60 * Sin 45^o = V_y = 42.43 m/s

Note: Remember that horizontal velocity is constant, but the vertical velocity we calculated above is only the initial vertical velocity.

From here we need to use the initial vertical velocity to find the time it takes the car to reach the top of its path and fall to the ground. Let’s think about this in two parts; the time it takes to reach h_MAX first:

t_TOP = 42.43/10 = 4.243 s

Now what about the time it takes to fall from the maximum height? Well first we need to know the maximum height:

h_TOP = ½ * 10 * 4.243² = 90.02 m

h_MAX = h_TOP + h_o =  90.02 + 315 = 405.02 m

Now we calculate the time it takes to fall from a height of 405.02 m:

405.02 = ½ * 10 * t_DOWN²

t_DOWN² = 81.004

t_DOWN = 9.000 s

Putting these two times together, we have the total time it takes the car to travel up to its maximum height and then fall back down. This is the total time in the air and this is the time we will want to use to solve for horizontal distance.

t_TOTAL = t_TOP + t_DOWN = 4.243 + 9.000 = 13.243 s

d_x = 42.43 * 13.243 = 561.9 m

You will want to make sure that the landing surface is centered 561.9 m from the base of the cliff.