Q71 / 90 · JEE Main 2024 (06 Apr Shift 2) · Physics

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$. Time required to reach the ground, if it is dropped from the top of the tower, is :

  • Correct answer
    A
    $\sqrt{t_1 t_2}$
  • B
    $\sqrt{t_1+t_2}$
  • C
    $\sqrt{t_1-t_2}$
  • D
    $\sqrt{\frac{t_1}{t_2}}$

Solution

$\begin{aligned} & \mathrm{t}_1=\frac{\mathrm{u}+\sqrt{\mathrm{u}^2+2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}_2=\frac{-\mathrm{u}+\sqrt{\mathrm{u}^2+2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}=\frac{\sqrt{2 \mathrm{gh}}}{\mathrm{g}} \\ & \mathrm{t}_1 \mathrm{t}_2=\frac{\left(\mathrm{u}^2+2 \mathrm{gh}\right)-\mathrm{u}^2}{\mathrm{~g}^2}=\frac{2 \mathrm{gh}}{\mathrm{g}^2}=\mathrm{t}^2 \\ & \Rightarrow \mathrm{t}=\sqrt{\mathrm{t}_1 \mathrm{t}_2}\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions