A ball thrown upward from the top of a tower with speed v reaches the ground in $t_{1}$ second. If this ball is thrown downward from the top of the same tower with speed v it reaches the ground $t_{2}$ second. In what time the ball shall reach the ground if it is allowed to falls freely under gravity from the top of the tower?

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$t_{1} + t_{2}$

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

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

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

answer is C.

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Detailed Solution

As, $h = - v t_{1} + \frac{1}{2} g t_{1}^{2}$

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

and $h = v t_{2} + \frac{1}{2} g t_{2}^{2}$

or $\frac{h}{t_{2}} = v + \frac{1}{2} g t_{2} . . . . . . . . . . . . . . . (i i)$

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

or $h = \frac{1}{2} g t_{1} t_{2}$

For free fall under gravity from the top of the tower

$h = \frac{1}{2} g t^{2}$

$∴ \frac{1}{2} g t_{1} t_{2} = \frac{1}{2} g t^{2}$

$\Rightarrow t = \sqrt{t_{1} t_{2}}$

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