Projection angle using max. height w/ proportions only

[amphiprion86]

Figures that the only problem I have trouble with is the one the book considers to be "easy":

Homework Statement

The speed of a projectile when it reaches its maximum height is one half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Homework Equations

h= (vi^2 sin^2 theta)/2g

The Attempt at a Solution

Complete brain fart--I don't even know where to start, beyond trying to divide h/2, which would be (vi^2 sin^2 theta)/4g, correct? That's about as far as I got. I just keep miring myself the further into it I go.

 

[Kurdt]

Speed of course is the magnitude of velocity. At its maximum height a projectile only has one component of velocity. Is that enough to get you started?

 

[Philosophaie]

Try breaking velocity into x and y components @ 1/2 max height and then finding speed from there the "sqrt of the sum of the squares".