JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 18)
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
20 J
5 J
15 J
zero
Explanation
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$$\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} \mathrm{mv}^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J} \end{aligned}$$
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