Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering if I was right about this simple problem:

A projectile is shot at an angle of 37.0° with the horizontal. The velocity of the projectile at its peak height is 16 m/s.

Then I was told to find the velocity in x direction.

So I wrote: Given that at its peak, the velocity of the projectile in y direction is equal to 0, but stated that at its peak, the velocity of the projectile is 16 m/s, this must be the velocity in x direction. So my final answer was: 16 m/s.

But then, my teacher just said it was wrong. That I had to find Vx using the formula Vx= Vcos(37°). I told her that 16 m/s could not be the initial velocity, since the problem says that 16 m/s is the velocity at its peak, therefore it must be Vx.

Am I right? If so, can you tell me a good explanation besides the one I said to prove my teacher wrong. If not, what am I missing?

share|cite|improve this question

Assuming the only acceleration is due to gravity in the $-y$ direction, you have reasoned correctly.

The $x$ component of velocity is constant and the $y$ component of velocity is zero at peak height.

The angle of the velocity vector is less than 37 degrees once the projectile is shot. You would only use 37 degrees to calculate the initial $y$ component of velocity.

share|cite|improve this answer
As the question is stated, I completely agree with this answer. It sounds like the teacher didn't write the question she intended to write, or it has been relayed to us incorrectly. Of course, she could simply be wrong. – Colin K Oct 23 '12 at 3:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.