Question

A body is projected such that the k.e at the topmost position is half of the initial k.e.what is its angle of projection with the horizontal

Answer 1

Given is that the kinetic energy at top is half of initial K.E .
Forming equation
1/2 mv^2cos^2x = (1/2)1/2mv^2
Eliminating 1/2 mv^2 from both the sides we get cos^2x = 1/2
Cos x =1/√2
This gives x=45°
I hope u got it .