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(B) At the maximum height of a projectile,the vertical component of velocity is zero. Therefore,the speed of the projectile at the maximum height is e

Vedclass · Accepted Answer

$\frac{3u^2}{8g}$

Vedclass · Answer

(B) At the maximum height of a projectile,the vertical component of velocity is zero. Therefore,the speed of the projectile at the maximum height is equal to its horizontal component of velocity,which is $v = u \cos 	heta$. According to the problem,the speed at maximum height is half the initial speed: $u \cos 	heta = \frac{u}{2} \Rightarrow \cos 	heta = \frac{1}{2}$. This implies $	heta = 60^{\circ}$. The formula for maximum height $H_{\max}$ is given by: $H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$. Substituting $	heta = 60^{\circ}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$: $H_{\max} = \frac{u^2 (\sqrt{3}/2)^2}{2g} = \frac{u^2 (3/4)}{2g} = \frac{3u^2}{8g}$.

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