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

Question

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

Professor PC Thomas Classes · Answer

At highest point, $v = u\cos(60^{\circ}) = u/2$. Kinetic energy $K' = \frac{1}{2}m(u/2)^2 = K/4$.