Kinematics

JEE Main level, numerical answer, medium

214 of 308

Question 214 of 308

The maximum height reached by a projectile is 64 m64 \mathrm{~m}. If the initial velocity is halved, the new maximum height of the projectile is _____ m\mathrm{m}.

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Answer: 16

Hmax⁡=u2sin⁡2θ2 gH1 max H2max⁡=u12u2264H2max⁡=u2(u/2)2H2max⁡=16 m\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}

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