# Height when a ball is thrown vertically at half its velocity

1. Jan 31, 2014

### Ace.

1. The problem statement, all variables and given/known data
A baseball is thrown vertically into the air with a velocity, v, and reaches a maximum height, h. At what height was the baseball with one-half its original velocity? Assume no air resistance.

2. Relevant equations
v22 = v12 + 2ad ?

3. The attempt at a solution
let 10m/s be the original velocity
d = $\frac{v^{2}_{2} - v^{1}_{2}}{2g}$
d = $\frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}$
d = 5.1

at 5 m/s:

d = $\frac{0 m/s - (5 m/s)^{2}}{2(-9.8 m/s^{2}}$
d = 1.28

$\frac{1.28}{5.1}$ = 0.25

but the answer is 0.75 of the original height. How to solve this?

2. Jan 31, 2014

### Dick

The d you are calculating is the distance between the point where the velocity v=0 and the point where the velocity is v/2. The point where v=0 is at the TOP of your trajectory.

3. Jan 31, 2014

### Ace.

I didn't fully understand what you mean

4. Feb 1, 2014

### Staff: Mentor

Ace, in the second part are not using the correct reference level.

The question you should be answering is: if a ball is thrown vertically upwards, at what height above the ground has its velocity dropped back to half of what it initially had?

Last edited: Feb 1, 2014
5. Feb 1, 2014

### Dick

You are calculating distances from the point where v=0. So the second distance you calculated isn't the distance from the ground. It's distance from the point where v=0. v=0 is at the top of the trajectory.