What is the initial projection angle of the projectile?

[batman11692003]

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.

 

[andrevdh]

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.

 

[batman11692003]

Thank you very much

 

[phiphi]

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