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

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The discussion centers on determining the initial projection angle of a projectile at its maximum height, given that the speed at maximum height is half the speed at half maximum height. The equations of motion for projectile motion are referenced, with specific focus on vertical and horizontal components of velocity. A calculation is presented that leads to the conclusion that the angle is approximately 67.79 degrees, derived from the relationship between vertical and horizontal velocities. The original poster seeks a detailed explanation of the calculations to better understand the solution process. Clarification of the steps involved in the derivation is requested to enhance comprehension of the topic.
militant07
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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 explination


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 :)?
 
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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.
 
anyone?
 

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