Kinematics

JEE Main level, single correct, medium

155 of 308

Question 155 of 308

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1t_1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2t_2. Time required to reach the ground, if it is dropped from the top of the tower, is :

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Answer: (A) t1t2\sqrt{t_1 t_2}

t1=u+u2+2ghgt2=−u+u2+2ghgt=2ghgt1t2=(u2+2gh)−u2 g2=2ghg2=t2⇒t=t1t2\begin{aligned} & \mathrm{t}_1=\frac{\mathrm{u}+\sqrt{\mathrm{u}^2+2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}_2=\frac{-\mathrm{u}+\sqrt{\mathrm{u}^2+2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}=\frac{\sqrt{2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}_1 \mathrm{t}_2=\frac{\left(\mathrm{u}^2+2 \mathrm{gh}\right)-\mathrm{u}^2}{\mathrm{~g}^2}=\frac{2 \mathrm{gh}}{\mathrm{g}^2}=\mathrm{t}^2 \\ & \Rightarrow \mathrm{t}=\sqrt{\mathrm{t}_1 \mathrm{t}_2}\end{aligned}

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