Q 11-03-052JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A ball of mass $100\ \text{g}$ is projected with velocity $20\ \text{m/s}$ at $60^\circ$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is
Answer: (B) $15\ \text{J}$
At the highest point only the vertical component of velocity is lost; the horizontal component stays.
$$\Delta K = \frac{1}{2}m(u\sin60^\circ)^2 = \frac{1}{2}(0.1)\left(20\times\frac{\sqrt3}{2}\right)^2 = \frac{1}{2}(0.1)(300) = 15\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics