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":

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{-}}{\mathit{g}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{-}}{\mathbf{2}}{\mathit{g}}{\mathbf{\u2206}}{\mathit{y}}}$

The maximum height reached:

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

At maximum height, the speed of the projectile is equal to v_{x} since v_{y} = 0.

The speed at maximum height, v_{hmax} = v_{0}cosθ

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?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Projectile Motion: Positive Launch concept. If you need more Projectile Motion: Positive Launch practice, you can also practice Projectile Motion: Positive Launch practice problems.

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Based on our data, we think this problem is relevant for Professor Voronine's class at USF.