JEE Main 2025 Jan 22 Shift 2, Physics Q16: Work Energy Theorem And Its Applications

At the highest point of a projectile, only the horizontal component of velocity survives: $v_x=v\cos 60^{\circ }=v/2$.

Initial KE: $K_i=\frac{1}{2}m{v}^2$.

KE at highest point: $K_f=\frac{1}{2}m(v\cos 60^{\circ }{)}^2=\frac{1}{2}m\cdot \frac{v^2}{4}=\frac{m{v}^2}{8}$.

Decrease in KE: $\Delta K=K_i-K_f=\frac{m{v}^2}{2}-\frac{m{v}^2}{8}=\frac{3m{v}^2}{8}$.

Substituting $m=0.1\text{kg}$, $v=20\text{m/s}$:

$\Delta K=\frac{3\times 0.1\times 400}{8}=\frac{120}{8}=15\text{J}$.