Projectile Motion Problems: Questions and Answers

Assuming that we all already have some knowledge of projectile motion, we’ll not be going over much of the details involved but to focus mainly on problems involving projectile motion. If you need to read some more on projectile motion then please click here.

In previous articles we said that a body that is in projectile motion will move in two dimensions simultaneously. These bodies are resolved into their perpendicular components which includes the vertical component Vy and the horizontal component Vx. We also said that they must be dealt with simultaneously. A body in projectile motion will look somewhat like the diagram below:


To be able to solve projectile motion problems you first need to know how to resolve vectors, to read more about resolving vectors click here.

Below are some formulas associated with projectile motion, we’ll be using them as we work through the questions below. Remember that objects that possess projectile motion moves in two dimensions simultaneously (horizontal and vertical), therefore we will have formulas for both the vertical component as well as the horizontal component. These are treated separately.

Projectile Formulas for Vertical Motion:

Uy=U×sinθ
Vy = Uy + at

Where:
  • Uy represents the initial velocity of the vertical component.
  • U represents initial velocity
  • θ represents the angle formed with the horizontal.
  • Vy represents the final velocity of the vertical component.
  • a and t represents acceleration and time respectively.

Projectile Formulas for Horizontal Motion:

Ux=U×cosθ
Vx = Ux + at

Where:
  • Ux represents the initial velocity of the horizontal component.
  • U represents initial velocity
  • θ represents the angle formed with the horizontal.
  • Vx represents the final velocity of the horizontal component.
  • a and t represents acceleration and time respectively.

Note: For objects moving with projectile motion in the horizontal direction you will find that initial velocity of the horizontal component and the final velocity of the horizontal component are equal this is because there is no acceleration taking place along the horizontal.

The equations stated above for both vertical and horizontal isn’t the complete list of equations that can be used but these are the ones we will be using during our calculations here. If you look at the second equation in both cases you will realize that those equations are similar to the equations of motion. Therefore the equations that are not stated above are also equations derived from the equations of motions, all you need to do is to replace the velocities with these involved in projectile motion.

Other general Formulas:







Where:

  • V represents velocity. Use this formula in cases where they ask for the velocity or magnitude.
  • θ is the angle
  • The last formula is usually used to find the direction.
Problem # 1

A particle is projected with a speed of 25m/s at 300 above the horizontal:

    a) Find the time taken to reach the highest point of the trajectory.
    b) Find the magnitude and direction of the velocity after 2 seconds.


Answers


     a) The first part of the question asked for the time taken to reach the highest point. To make our calculations easier we’ll work out our answer through steps.

Step 1
Write down all the variables that are given. This way you will have easier access to them when you’re ready to use them in calculations also this helps you to choose the best possible formula for the question.
  • Θ = 300
  • U = 25m/s
  • Uy which is the initial velocity is = ???
  • a = -g = -10m/s2, this is because the particle is moving upward against gravity.
  • Vy = 0. In this part of the question the final velocity is at the highest point and at the highest point the particle stops moving in the vertical direction therefore the final velocity of the vertical component, Vy, is zero.
  • t =???


Step 2
The next step would be to choose the best possible formula to solve the problem. Looking at the variables we have and what we want to find (time) the best possible formula would then be:

                Vy = Uy + at

Where a is acceleration and t is time.

We already know the values for most of the variables above except the time, t, which we are trying to find and the value of one other variable that is needed to solve for time, this variable is Uy, which is the initial velocity.

To solve for the initial velocity, Uy, we simply use the formula:

Uy=U×sinθ

After filling in the values in the above equation and solve for Uy your answer should know look something like this:

Uy=U×sinθ
Uy=25×sin300
Uy=12.5m/s


Step 3
Now that we have all the values we need as well as the formula, we can now go ahead and solve for time (t):

Transpose to make time the subject


   b) This part of the question asked for the magnitude and the direction of the velocity after 2 seconds. To make our calculations allot easier we will split this question into two. First we will find the magnitude and then we’ll find the direction.

To find the magnitude we will use the formula:


If you look at the equation you’ll realize that we can’t solve for the magnitude right away because we don’t have the values for Vx or Vy. We can’t use the value we got for Vy in part (a) of the question here because that velocity was the finally velocity of the vertical when the object reached maximum height(which took 1.25seconds) but in this case we have already passed the maximum height so there’s a new final velocity for the vertical which is after 2 seconds. The initial velocity for the vertical however will remain the same.


Finding Final Velocity for Vertical, Vy:
Uy = 12.5m/s
a = -10
t = 2 sec
Vy =?

Using the formula Vy = Uy + at:

Vy = Uy + at
     = 12.5 + -10 x 2
     = -7.5m/s

Finding final velocity for Horizontal, Vx:
Remember that the Initial and final velocities for the horizontal are the same, so you can use the formula below to find the initial velocity of the horizontal, Ux, which will also be the answer for the final velocity of the horizontal, Vx:

Ux = U×cosθ
     = 25 x cos 30
     = 21.65m/s
Therefore Ux = Vx = 21.65m/s

Now that we have found both Vy and Vx we can go ahead and solve for the magnitude:


Now for the second part of the question, we must find the direction of the velocity after 2 seconds. We can use the following formula to find the direction:

Just substitute the values for Vy and Vx into the equation and then solve for the angle.

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8 Response to Projectile Motion Problems: Questions and Answers

September 20, 2011 at 4:43 AM

Really very interesting post!

September 29, 2011 at 5:27 AM

I really enjoy reading your posts. They help also my students to understand physical matters better.

Anonymous
January 18, 2012 at 12:47 AM

This post was very helpful.

Anonymous
February 14, 2012 at 2:55 AM

Thank you helped me out a lot

Anonymous
June 5, 2012 at 1:10 PM

thanks,but please add more questions...

Anonymous
September 15, 2012 at 2:04 AM

thats awesome. now my problems are solved

Anonymous
June 21, 2013 at 5:19 AM

I revised the lessons better than I packed on my mind

Anonymous
February 12, 2014 at 3:29 AM

you throw a ball downward at a speed of 2 m/s.how fast is it moving when it hits the sidewalk 2.5 below/

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