**Problem**
When a ball reaches its maximum height, its speed is one-fourth the speed with which it was thrown. What was the launch angle?

**Concepts that I understand**
I understand that at the maximum height, the velocity of the ball only has a component in the x-direction, and I've made the distinction that the x-component of the velocity must be 1/4 the launch velocity, Vo. I did some "guess and check" work and got that the angle would be about 75.5, but I have no idea how to prove that mathematically.

**Attempt at a solution**
$\displaystyle Vix=\sqrt{Vix^2+viy^2}/4$

$\displaystyle 4Vix=\sqrt{Vix^2+viy^2}$

$\displaystyle 16Vix^2=Vix^2+Viy^2$

$\displaystyle 15Vix^2=Viy^2$

$\displaystyle Arctan(15/4)=75.1 degrees$