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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2025
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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

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(B) 15 J

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\(\begin{aligned} & \mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{k}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=\frac{1}{8} m v^2 \\ & \Delta \mathrm{k}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{3}{8} \mathrm{mv}^2=\frac{3}{8} \times 0.1 \times 400=15 \mathrm{~J}\end{aligned}\)
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