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

Correct answer: 16. H_max =(u^2 sin^2 θ)/(2 ~g) fracH_1 max H_2 max =(u_1^2)/(u_2^2) frac64H_2 max =(u^2)/((u / 2)^2) H_2 max =16 ~m

Q86 / 90 · JEE Main 2024 (05 Apr Shift 2) · Physics

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 _____ $\mathrm{m}$.

Correct answer: 16

Solution

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

Hence, the correct answer is 16.

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