What is the initial projection angle of the projectile?

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The discussion centers on determining the initial projection angle of a projectile, given that its speed at maximum height is half of its speed at half maximum height. The relevant formula used is v^2 = u^2 + 2as, emphasizing that at the peak, the vertical speed component is zero, leaving only the horizontal component. The calculations involve the relationships between initial speed, height, and the angle of projection, leading to the conclusion that sin^2(θ) equals 6/7. This results in an initial projection angle of approximately 67.8 degrees. The analysis effectively combines kinematic equations to solve for the angle.
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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? Please help. Thanks.
 
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You need only one formula

v^2 = u^2 +2as

to solve this problem. The other point that you need to realize is that at the top of its trajectory the vertical component of its speed is zero. This means that at the top the speed consists of only the horizontal component

v_o \cos(\theta _o)

of the projectile.
 
Thank you very much
 
v:the speed at maximum height
h:half of maximum height
Vi: initial speed
at maximum height: 2h=vi^2sin^2(Ѳ)/2g
v=vicos(Ѳ)
at half of maximum height: h=visin(Ѳ)t-0.5gt^2
4v^2=Vx^2+Vy^2
Vx=Vicos(Ѳ)
Vy=visin(Ѳ)-gt
==sin^2(Ѳ)=6/7=>Ѳ=67.8 degree
 
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