What Is the Initial Projection Angle of a Projectile at Maximum Height?

[militant07]

1. Homework Statement
The speed of projectile when it reaches its maximum height is one it half speed when it’s half maximum height. What is initial projection angle of the projectile?


2. Homework Equations
I know it has been asked several times but no one give the answer with explanation

The Attempt at a Solution

I tried with h=viy^2/2g but I don’t how to start.

Can anyone please solve it with explanation? pretty please :)?

 

[militant07]

vx = vcos(theta)
vy = vsin(theta)
vyh (vy halfway up) = sqrt(0.5)*vy
vh (v halfway up) = sqrt(vx^2+vyh^2) = sqrt(vx^2+0.5vy^2)
vm (v at max height) = vx
vx = vh/2 = sqrt(vx^2+0.5vy^2)/2
vx^2 = (vx^2+0.5vy^2)/4
0.75vx^2 = 0.125vy^2
vy/vx = sqrt(6)
theta = arctan(sqrt(6)) = 67.7923457 deg

that what i found from http://answers.yahoo.com/question/index?qid=20100919112019AAwQ3Of

could somebody please explain everything he done.

 

[militant07]

anyone?