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

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Solution

Q.
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 [2024]

1 $\sqrt{t_{1} + t_{2}}$

2 $\sqrt{t_{1} - t_{2}}$

3 $\sqrt{t_{2} t_{1}}$

4 $\sqrt{\frac{t_{1}}{t_{2}}}$

Ans.
(3)

Using: $s = u t + \frac{1}{2} a t^{2}$

In situation (C): $h = \frac{1}{2} g t_{3}^{2}$ ...(i)

Situation (A): $- h = u t_{1} - \frac{1}{2} g t_{1}^{2}$

$h = - u t_{1} + \frac{1}{2} g t_{1}^{2}$ ...(ii)

Situation (B): $- h = - u t_{2} - \frac{1}{2} g t_{2}^{2}$

$h = u t_{2} + \frac{1}{2} g t_{2}^{2}$ ...(iii)

From equation (ii) and equation (iii): $eq (ii) \times t_{2} + eq (iii) \times t_{1}$

$h t_{2} + h t_{1} = \frac{1}{2} g t_{1}^{2} t_{2} + \frac{1}{2} g t_{2}^{2} t_{1} \Rightarrow h (t_{2} + t_{1}) = \frac{1}{2} g t_{1} t_{2} (t_{2} + t_{1})$

$\Rightarrow h = \frac{1}{2} g t_{1} t_{2}$ ...(iv)

Using (i) and (iv), $\frac{1}{2} g t_{3}^{2} = \frac{1}{2} g t_{1} t_{2} \Rightarrow t_{3} = \sqrt{t_{1} \cdot t_{2}}$

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