Projection angle using max. height w/ proportions only

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SUMMARY

The discussion focuses on determining the initial projection angle of a projectile based on its speed at maximum height and half of that height. The key equation involved is h = (vi^2 sin^2 theta) / 2g, where h is the maximum height, vi is the initial velocity, theta is the projection angle, and g is the acceleration due to gravity. The problem states that the speed at maximum height is half of the speed at half the maximum height, leading to the need for a breakdown of velocity components to solve for theta. Participants suggest analyzing the velocity components at half the maximum height to derive the necessary relationships.

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amphiprion86
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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.
 
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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?
 
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".
 

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