Question

A particle is projected at ${60}^{\circ }$ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is:

Answer 1

Let initial vel, be $v_0$
Initial kinetic energy, $k=\frac{1}{2}m{v}_0^2$
At highest point v $=v_{0}cos60=\frac{v_0}{2}$
So KE at highest point $=\frac{1}{2}m{v}^2=\frac{1}{2}m\frac{v_0^2}{4}=\frac{K}{4}$