A body is projected vertically downwards from A, the top of the tower, reaches the ground in $t_{1}$ seconds. If it is projected upwards with same velocity, it reaches the top in $t_{2}$ seconds. At what time will it reach the ground if it is dropped from A?

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$t_{1} t_{2}$

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

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

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

answer is C.

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Detailed Solution

For downward motion,

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

For upward motion,

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

when dropped from A

$h = \frac{1}{2} {gt}^{2} . . . . . . . . (iii)$

From (i) & (ii)

${ut}_{1} + \frac{1}{2} {gt}_{1}^{2} = - {ut}_{2} + \frac{1}{2} {gt}_{2}^{2}$

$\Rightarrow u (t_{1} + t_{2}) = \frac{1}{2} g (t_{2}^{2} - t_{1}^{2}) \Rightarrow u = \frac{1}{2} g (t$ ₂- t₁)

Substituting in equation (i),

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

for dropped body,

$∴ R e q u i r e d t i m e t = \sqrt{\frac{2 h}{g}} = \sqrt{t_{1} t_{2}}$

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