The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new maximum height of the projectile is _____ $\mathrm{m}$.

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}} \ & \frac{\mathrm{H}_{1 	ext { max }}}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}_1^2}{\mathrm{u}_2^2} \ & \frac{64}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}^2}{(\mathrm{u} / 2)^2} \ & \mathrm{H}_{2 \max }=16 \mathrm{~m}\end{aligned}$

The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new…

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