Skip to main content

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?
Vx = 3.5 m/s                            Vy = 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 = dy = ½ * 10 * t2 = 300
300 * 2 / 10 = t2 = 60
t = 7.75 s

d)     How far from the cliff will this bird land?
dx = 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]
Vy = 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]
Vx = 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) = 87.41o 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 = dy = 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 32o and an angle of 58o if the bean bag is thrown with an initial velocity of 6 m/s:
                  32o                                                                   58o
i.        What are the components of velocity?
            Vx32: Cos 32o = Vx32/6                                                Vx58: Cos 58o = Vx58/6           
            6 * Cos 32o = Vx32 = 5.09 m/s                                    6 * Cos 58o = Vx58 = 3.18 m/s

            Vy32: Sin 32o = Vy32/6                                                 Vy58: Sin 58o = Vy58/6
            6 * Sin 32o = Vy32 = 3.18 m/s                                     6 * Sin 58o = Vy58 = 5.09 m/s

ii.      What is the maximum height of the bean bag’s motion?
            tTOP32 = Vy/10 = 3.18/10 = 0.318s                   tTOP58 = Vy58/10 = 5.09/10 = 0.509s
            hMAX32 = ½ * 10 * 0.3182 = 0.506 m               hMAX58 = ½ * 10 * 0.5092 = 1.295 m

iii.    How long will the bean bag be in the air?
            tTOTAL32 = 2 * tTOP32 = 0.636 s                        tTOTAL58 = 2 * tTOP58 = 1.018 s

iv.    How far away from you will the bag land?
            dx32 = 5.09 * 0.636 =   3.24 m                         dx58 = 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 45o 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:

            Vx: Cos 45o = Vx/60                                       Vy: Sin 45o = Vy/60
            60 * Cos 45o = Vx = 42.43 m/s                       60 * Sin 45o = Vy = 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 hMAX first:

tTOP = 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:

hTOP = ½ * 10 * 4.2432 = 90.02 m

hMAX = hTOP + ho =  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 * tDOWN2

tDOWN2 = 81.004

tDOWN = 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.

tTOTAL = tTOP + tDOWN = 4.243 + 9.000 = 13.243 s

dx = 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.


Popular posts from this blog



In  this  configuration  the  input  is  applied  between the  base  and  the  collector and  the  output  is  taken  from  the  collector  and  the  emitter.  Here  the  collector  is common to both the input and the output circuits as shown in Fig.

  Common Collector Transistor Circuit

In  common  collector  configuration  the  input  current  is  the  base current  IB  and  the output current is the emitter current IE. The ratio of change in emitter current to the  change in the base current is called current amplification factor.

It is represented by


A test  circuit  for determining the  static characteristic  of an NPN transistor is shown in Fig. In this circuit the collector is common to both the input and the output circuits.   To   measure   the   base   and   the   emitter   currents,   milli   ammeters   are connected in series with the base and the emitter circuits. Voltmeters are connected   across the input an…

XII - Ch# 12 : Electrostatics :Solved Numericals

Solution Manual : Mathematical methods for physicists 5th edition Arfken and Weber

DJ VU Reader
Book Description Now in its 7th edition, Mathematical Methods for Physicists continues to provide all the mathematical methods that aspiring scientists and engineers are likely to encounter as students and beginning researchers. This bestselling text provides mathematical relations and their proofs essential to the study of physics and related fields. While retaining the key features of the 6th edition, the new edition provides a more careful balance of explanation, theory, and examples. Taking a problem-solving-skills approach to incorporating theorems with applications, the book's improved focus will help students succeed throughout their academic careers and well into their professions. Some notable enhancements include more refined and focused content in important topics, improved organization, updated notations, extensive explanations and intuitive exercise sets, a wider range of problem solutions, improvement in the placement, and a wider ra…