Question

a shot is fired at an angle of 60 degree to the horizontal with kinetic energy E. if air resistance is ignored ,the kinetic energy at the top of the trajectory is

Answer 1

Well ... You must be knowing that the velocity of a body at the top of trajectory remains only the horizontal component of velocity with which it was fired with. Suppose a body is fired at angle x with horizontal . So it's velocity at top of trajectory will be equal to ( v cos x )
We know the general formula of kinetic energy I.e . 1/2 mv^2
So kinetic energy at top will be 1/2m (v cos x )^2 that is equal to 1/2mv^2cos^2x .
I hope it helped.

Answer 2

Ans. is E/4 ... solution is in the pic ... :)