Q61 / 75 · JEE Main 2025 (22 Jan Shift 2) · Physics
A ball of mass 100 g is projected with velocity $20 \mathrm{~m} / \mathrm{s}$ at $60^{\circ}$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is
- A5 J
- Correct answerB15 J
- C20 J
- Dzero
Solution

$\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} m v^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J}\end{aligned}$
Hence, the correct answer is option: 'B'.