Q 11-03-054JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
A ball having kinetic energy $KE$ is projected at an angle of $60^\circ$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?
Answer: (D) $\dfrac{KE}{4}$
At the highest point only the horizontal component $u\cos60^\circ = u/2$ remains.
$$KE_{top} = \frac{1}{2}m\left(\frac{u}{2}\right)^2 = \frac{1}{4}\cdot\frac{1}{2}mu^2 = \frac{KE}{4}$$
Solution by Sreeraj P, M.Sc Physics