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(B) Let the initial velocity be $u$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = 120 \ m$.Since $\sin 45^{\circ} = \fra

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let the initial velocity be $u$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = 120 \ m$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, we have $H = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g} = 120 \ m$, so $u^2 = 480g$.The kinetic energy before the bounce is $K_i = \frac{1}{2}mu^2$.After the bounce, the kinetic energy is $K_f = \frac{1}{2}K_i = \frac{1}{4}mu^2 = \frac{1}{2}mv^2$, where $v$ is the velocity after the bounce.Thus, $v^2 = \frac{u^2}{2} = \frac{480g}{2} = 240g$.The new maximum height $h$ is given by $h = \frac{v^2 \sin^2 30^{\circ}}{2g}$.Substituting $v^2 = 240g$ and $\sin 30^{\circ} = \frac{1}{2}$, we get $h = \frac{240g 	imes (1/2)^2}{2g} = \frac{240g 	imes (1/4)}{2g} = \frac{60}{2} = 30 \ m$.

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