If the kinetic energy at maximum height is 2/5 the kinetic energy at half the maximum height, find the angle of projection.
In other words: K.E. at H(max) = 0.4*K.E. at 0.5*H(max)
I got 60 degrees, however, my teacher told me that the answer is 30 degrees. Please solve it for me so I can see where I went wrong.
Well here's my edit:
Consider u to be the velocity at H(max) Consider v to be the velocity at half of H(max) Consider c to be initial velocity given Consider @ to be angle projected at sqr() means square root
u^2 = 0.4v^2
u is obviously c*cos(@) v I found to be: sqr((0.5u^2)(1+(cos(@))^2)
Using this, I substituted into the previous equation and solved to attain: cos@ = 0.5
So.... That gives me 60 degrees. Am I right? Or is there some sort of catch?