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 _______.
- A \(\sqrt{t_1 t_2}\)
- B \(\sqrt{t_1-t_2}\)
- C \(\sqrt{\frac{t_1}{t_2}}\)
- D \(\sqrt{t_1+t_2}\)
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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