Height when a ball is thrown vertically at half its velocity

[itex][/itex]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 = [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?