Projectile Motion Example Problem – Physics Homework Help 3


Throwing or shooting a projectile follows a parabolic course. If you know the initial velocity and angle of elevation of the projectile, you can find its time aloft, maximum height or range. You can also its altitude and distance travelled if given a time. This example problem shows how to do all of these.

Projectile Motion Example Problem:
A cannon is fired with muzzle velocity of 150 m/s at an angle of elevation = 45°. Gravity = 9.8 m/s2.
a) What is the maximum height the projectile reaches?
b) What is the total time aloft?
c) How far away did the projectile land? (Range)
d) Where is the projectile at 10 seconds after firing?

Projectile motion problem setup illustrationLet’s set up what we know. First, let’s define our variables.

V0 = initial velocity = muzzle velocity = 150 m/s
vx = horizontal velocity component
vy = vertical velocity component
θ = angle of elevation = 45°
h = maximum height
R = range
x = horizontal position at t=10 s
y = vertical position at t=10 s
m = mass of projectile
g = acceleration due to gravity = 9.8 m/s2

Part a) Find h.

The formulas we will be using are:

d = v0t + ½at2

and

vf – v0 = at

In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth.

The initial vertical velocity is
v0y = v0·sinθ
v0y = 150 m/s · sin(45°)
v0y = 106.1 m/s

Now we know the beginning and final velocity. The next thing we need is the acceleration.

The only force acting on the projectile is the force of gravity. Gravity has a magnitude of g and a direction in the negative y direction.

F = ma = -mg

solve for a

a = -g

Now we have enough information to find the time. We know the initial vertical velocity (V0y) and the final vertical velocity at h (vhy = 0)

vhy – v0y = at
0 – v0y = -9.8 m/s2·t
0 – 106.1 m/s = -9.8 m/s2·t

Solve for t

projectile motion math step 3

t = 10.8 s

Now solve the first equation for h

h = v0yt + ½at2
h = (106.1 m/s)(10.8 s) + ½(-9.8 m/s2)(10.8 s)2
h = 1145.9 m – 571.5 m
h = 574.4 m

The highest height the projectile reaches is 574.4 meters.

Part b: Find total time aloft.

We’ve already done most of the work to get this part of the question if you stop to think. The projectile’s trip can be broken into two parts: going up and coming down.

ttotal = tup + tdown

The same acceleration force acts on the projectile in both directions. The time down takes the same amount of time it took to go up.

tup = tdown

or

ttotal = 2 tup

we found tup in Part a of the problem: 10.8 seconds

ttotal = 2 (10.8 s)
ttotal = 21.6 s

The total time aloft for the projectile is 21.6 seconds.

Part c: Find range R

To find the range, we need to know the initial velocity in the x direction.

v0x = v0cosθ
v0x = 150 m/s·cos(45)
v0x = 106.1 m/s

To find the range R, use the equation:

R = v0xt + ½at2

There is no force acting along the x-axis. This means the acceleration in the x-direction is zero. The equation of motion is reduced to:

R = v0xt + ½(0)t2
R = v0xt

The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.

R = 106.1 m/s · 21.6s
R = 2291.8 m

The projectile landed 2291.8 meters from the canon.

Part d: Find the position at t = 10 seconds.

The position has two components: horizontal and vertical position. The horizontal position, x, is far downrange the projectile is after firing and the vertical component is the current altitude, y, of the projectile.

To find these positions, we will use the same equation:

d = v0t + ½at2

First, let’s do the horizontal position. There is no acceleration in the horizontal direction so the second half of the equation is zero, just like in Part c.

x = v0xt

We are given t = 10 seconds. V0x was calculated in Part c of the problem.

x = 106.1 m/s · 10 s
x = 1061 m

Now do the same thing for the vertical position.

y = v0yt + ½at2

We saw in Part b that v0y = 109.6 m/s and a = -g = -9.8 m/s2. At t = 10 s:

y = 106.1 m/s · 10 s + ½(-9.8 m/s2)(10 s)2
y = 1061 – 490 m
y = 571 m

At t=10 seconds, the projectile is at (1061 m, 571 m) or 1061 m downrange and at an altitude of 571 meters.

If you need to know the velocity of the projectile at a specific time, you can use the formula

v – v0 = at

and solve for v. Just remember velocity is a vector and will have both x and y components.

This specific example can be easily adapted for any initial velocity and any angle of elevation. If the cannon is fired on another planet with a different force of gravity, just change the value of g accordingly.


About Todd Helmenstine

Todd Helmenstine is the physicist/mathematician who creates most of the images and PDF files found on sciencenotes.org. Nearly all of the graphics are created in Adobe Illustrator, Fireworks and Photoshop. Todd also writes many of the example problems and general news articles found on the site.


3 thoughts on “Projectile Motion Example Problem – Physics Homework Help

  • Wolf Suit

    Yeah. That 109.6? Pretty sure that’s not correct.

    sin(45) = 0.707106781186547
    0.707106781186547*150m/s = 106.0660171779821m/s

    Unless my calculator is broken, that’s the answer. I tried rounding to different sig digs and still could not for the life of me come up with 109.6m/s. I checked Pythagorean’s theorem and it doesn’t hold up that way either.

    (109.6^2)+(109.6^2)=C^2
    24,024.32=C^2
    C=154.9978064360912

    Again, I know I’m being ridiculous with the significant digits, but this is a fair difference in answers. I’m not trying to be argumentative either, I’m just wondering if I’ve made a mistake somewhere.

    • Todd Helmenstine Post author

      You are the one who is correct. I’m not really sure where I punched in my values, but sin(45) = cos(45) = 0.707. Use that with the velocity and get 106.1 m/s…NOT 109.6.

      I fixed the math in all the steps. It was kind of strange, but the velocity I had for the time aloft was the correct value, not the 109.6 I used everywhere else. Goes to show you should always check your work.

      Thanks for pointing out my error! 🙂

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