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Ace.

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## Homework Statement

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.

## Homework Equations

v

_{2}

^{2}= v

_{1}

^{2}+ 2ad ?

## The Attempt at a Solution

let 10m/s be the original velocity

d = [itex]\frac{v^{2}_{2} - v^{1}_{2}}{2g}[/itex]

d = [itex]\frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]

d = 5.1

at 5 m/s:

d = [itex]\frac{0 m/s - (5 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]

d = 1.28

[itex]\frac{1.28}{5.1}[/itex] = 0.25

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