A body is projected vertically upwards from… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Mechanics

2021mcqhard

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $S.$

Official previous-year question

Held on 22 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\sqrt{\frac{{R}_{e}}{2g}}[{(1+\frac{h}{{R}_{e}})}^{\frac{3}{2}}-1]$
B
$\sqrt{\frac{2{R}_{e}}{g}}[{(1+\frac{h}{{R}_{e}})}^{\frac{3}{2}}-1]$
C
$\frac{1}{3}\sqrt{\frac{{R}_{e}}{2g}}[{(1+\frac{h}{{R}_{c}})}^{\frac{3}{2}}-1]$
D
$\frac{1}{3}\sqrt{\frac{2{R}_{c}}{g}}[{(1+\frac{h}{{R}_{e}})}^{\frac{3}{2}}-1]$

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Solution

JEE Main 2021 Physics, Gravitation — solution figure

Applying energy conservation from ( $(1)\text{to}(2)$ ,

$\frac{1}{2}m\cdot (\frac{2\mathrm{GM}}{{R}_{e}})-\frac{GMm}{{R}_{e}}=\frac{1}{2}m{v}^{2}-\frac{GMm}{R+r}$

$\Rightarrow \frac{1}{2}m{v}^{2}=\frac{GMm}{R+r}$

$\Rightarrow v=\sqrt{\frac{2GM}{R+r}}=\frac{dr}{dt}$

$\Rightarrow \sqrt{2GM}{\int }_{0}^{t}dt={\int }_{{R}_{e}}^{{R}_{e}+h}(\sqrt{R+r})dr$

$\sqrt{2GM}\cdot t=\frac{2}{3}{[(R+r{)}^{\frac{3}{2}}]}_{{R}_{e}}^{{R}_{e}+h}$

$t=\frac{2}{3}\sqrt{\frac{{R}_{e}^{3}}{2\mathrm{GM}}}[{(1+\frac{h}{{R}_{e}})}^{\frac{3}{2}}-1]$

$\frac{GM}{{R}_{e}^{2}}=g$

$t=\frac{1}{3}\sqrt{\frac{2{R}_{e}}{g}}[{(1+\frac{h}{{R}_{e}})}^{\frac{3}{2}}-1]$

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