Lesson 26: Projectiles Launched Horizontally

The study of projectile motion brings together a lot of what you have learned in the past few chapters.

The Wile E. Coyote Effect

I’m sure all of you have seen the cartoon where the Coyote is chasing after the Road Runner and runs off of the cliff. He hangs in mid air for a second, looks down, and then starts to fall.

A few years ago some researchers in the U.S. went to elementary schools, junior and senior high schools, and universities and asked them to look at the following:

"Ignoring air resistance, which of the following correctly shows
what an object would do if it rolled off a cliff?"

Figure 1

The breakdown of answers they got was almost exactly the same at all ages.

The correct answer is actually number 3, and if you think about it using the physics you've studied it makes sense.

That doesn't tell us anything about what is happening vertically , which is completely separate from what the object is doing horizontally.

Looking at this problem as what is happening horizontally and vertically, you should get the idea that this is just like the components that we were just working on a couple of lessons back.

If we look at the horizontal and vertical components of an objects velocity as it rolls off a cliff, we would get something that looks like this.

Figure 2

To understand how to actually figure out questions involving these situations, it’s probably best to look at an example.

Example 1: I throw a ball off the edge of a 15.0m tall cliff. I threw it horizontally at 8.0m/s.

  1. Determine how much time it takes to fall.
  2. Determine how far from the base of the cliff it hits the ground.
  3. Determine how fast it is moving vertically when it hits the ground.
  4. Determine what its total velocity is when it hits the ground.


We’re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers. It actually would take the exact same amount of time for the object to hit the ground if I just dropped it straight down from the edge of the cliff , so let’s just calculate the time to fall that way. Remember to think vertically…

d = vit + 1/2 at2 <- Initial vertical velocity is zero so...
d = 1/2 at2
t = sqr root (2d/a) = sqr root (2 x 15.0 / 9.81)
t = 1.75 s


Well, we know it was in the air for 1.75s (from the previous question), and it was moving at a constant speed of 8.0m/s in the x-direction the whole time, so…

v = d/t
d = v t = (8.0m/s)(1.75s)
d = 14m

It will move 14m horizontally, so it hits the ground 14m away from the base of the cliff.


It has been accelerating down the whole time. We know that gravity is causing this acceleration, and that it wasn’t moving vertically at the start, so we can figure out how fast it is going (vertically) when it hits the ground.

vf2 = vi2 + 2ad
vf = sqr root (vi2 + 2ad) = sqr root (0 + 2 x 9.81 x 15.0)
vf = 17 m/s

d) Combination!

Figure 3

It’s total velocity is found by adding the horizontal and final vertical components of the velocity to find the resultant.

c2 = a2 + b2
= (8.0m/s)2 + (17m/s)2
c = 19m/s

tanΘ = opp/adj
= (8.0m/s) / (17m/s)
Θ = 25°

The object is moving at 19m/s at an angle of 25° below the horizontal when it hits the ground.

Although there will always be slight differences in actual problems, this is the standard sort of questions that you will be asked for these types of questions.

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