# Problem: 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?

###### FREE Expert Solution

In this problem, we are going to express the information given using equations then solve.

Uniform accelerated motion (UAM) equations, a.k.a. "kinematics equations":

The maximum height reached:

$\overline{){{\mathbf{h}}}_{\mathbf{m}\mathbf{a}\mathbf{x}}{\mathbf{=}}\frac{{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbit{\theta }}{\mathbf{2}\mathbf{g}}}$

At maximum height, the speed of the projectile is equal to vx since vy = 0.

The speed at maximum height, vhmax = v0cosθ

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###### Problem Details

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?