An object of mass 5 kg is thrown vertically upwards from the ground… | JEE Main 2022 (24 Jun, Shift 2)

Q19JEE Main 2022 · 24 JUN · SHIFT 2Difficult

Inertia, First & Second Laws

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to: [Use g = 10 ms ${}^{-2}$ ].

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Answer: (b)

Up: $a_1 = g + \frac{f}{m} = 10 + 2 = 12\text{ m/s}^2$ ; down: $a_2 = g - \frac{f}{m} = 8\text{ m/s}^2$
Same height $h = \frac{1}{2}a t^2$ both ways, so $t \propto \frac{1}{\sqrt{a}}$ .
$\frac{t_a}{t_d} = \sqrt{\frac{a_2}{a_1}} = \sqrt{\frac{8}{12}} = \frac{\sqrt{2}}{\sqrt{3}}$
So the answer is (b) $\sqrt{2} : \sqrt{3}$ .

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