# Acceleration at its highest point?

1. Sep 8, 2009

### maegenr8

1. The problem statement, all variables and given/known data
I need to find the acceleration of the object at its highest point?? I thought it would be zero, but that is wrong.

-The initial velocity is 26 m/s at 54 degrees above the horizontal axis.
-it's initial horizontal velocity is 15.3 m/s and vertical is 21.0 m/s (is this acceleration?)
-It reaches a height of 22.6 meters
-in 2.15 seconds
-the balls velocity at its highest point is 15.3 m/s

What's the balls acceleration at its highest point?
For how long a time is the ball in the air?
When the ball lands on the court, how far is it from the place where it was hit?

Please explain. I'm not sure which values to use for initial velocity, etc. when I use the formulas.

2. Relevant equations

3. The attempt at a solution

I thought I should use the velocity-time equation. final velocity=initial velocity + acceleration*time
I did 15.3= 26+ 2.15a but the answer was incorrect.
What am I doing wrong?

2. Sep 8, 2009

### w3390

The main thing you need to do is keep you dimensions separate. Work out the equation for all motion in the x plane. Then work out the equation for all motion in the y plane. Try this out first.

3. Sep 8, 2009

### maegenr8

when working it for the values in the x plane would i just use 15.3 as initial velocity or the 26?

4. Sep 8, 2009

### w3390

You would use the 15.3 because that is the x component of the velocity.

5. Sep 8, 2009

### maegenr8

okay, 15.3 is the velocity at the balls highest point. is that because it's vertical velocity at its highest point is zero (because at that instance its not going up or down) but it is still moving horizontally and that's why it's 15.3?

6. Sep 8, 2009

### w3390

I'm a bit confused by what you are saying. You are correct that the vertical velocity of the ball at its highest point is zero and that the horizontal velocity at its highest point is 15.3, but you need to be more specific when stating the ball's velocity