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JEE Mains · Physics · STD 11 - 2. motion in straight line

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 _______.

  1. A \(\sqrt{t_1 t_2}\)
  2. B \(\sqrt{t_1-t_2}\)
  3. C \(\sqrt{\frac{t_1}{t_2}}\)
  4. D  \(\sqrt{t_1+t_2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{t_1 t_2}\)

Step-by-step Solution

Detailed explanation

\(t_1=\frac{u+\sqrt{u^2+2 g h}}{g}\) \(t_2=\frac{-u+\sqrt{u^2+2 g h}}{g}\) \(t=\frac{\sqrt{2 g h}}{g}\) \(t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2\) \(\Rightarrow t=\sqrt{t_1 t_2}\)
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