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Old Oct 14th 2017, 04:39 PM   #1
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Projectile Motion Derivation

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.
Projectile Motion Derivation-physics-problem.png
Attempt at a solution
$\displaystyle Vix=\sqrt{Vix^2+viy^2}/4$
$\displaystyle 4Vix=\sqrt{Vix^2+viy^2}$
$\displaystyle 16Vix^2=Vix^2+Viy^2$
$\displaystyle 15Vix^2=Viy^2$
$\displaystyle Arctan(15/4)=75.1 degrees$
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Old Oct 14th 2017, 07:45 PM   #2
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At its highest point, the projectile has velocity strictly in the horizontal direction, i.e. $v = v_x$ ...

$v_x = v_0 \cos{\theta} \implies v_0 \cos{\theta} = \dfrac{v_0}{4} \implies \cos{\theta} = \dfrac{1}{4} \implies \theta = \arccos\left(\dfrac{1}{4}\right) \approx 75.5^\circ = \arctan(\sqrt{15})$
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Old Oct 14th 2017, 08:17 PM   #3
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Thanks!

I had actually justtt figured this out when I got the notification. Thanks a lot for the answer though!
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