Physics · Motion in Plane

JEE Main 2024 — 5 April, Shift 2 — Question 57

The maximum height reached by a projectile is 64 m . If the initial velocity is halved, the new maximum height of the projectile is $\qquad$ m .

Answer: 16

Numerical answer — enter this value.

$\quad \mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

$\frac{\mathrm{H}_{1 \max }}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}_{1}^{2}}{\mathrm{u}_{2}^{2}}$

$\frac{64}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}^{2}}{(\mathrm{u} / 2)^{2}}$

$\mathrm{H}_{2 \text { max }}=16 \mathrm{~m}$

Answer key and solution verified before publishing.

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