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1st Equation of Motion

A ball thrown...

Question

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 in $t_{2}$ second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from the top of the tower?

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

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

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

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t_{1} + t_{2}

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Solution

The correct option is C $\sqrt{t_{1} t_{2}}$
$h = - v t_{1} + \frac{1}{2} g t_{1}^{2}, h = v t_{2} + \frac{1}{2} g t_{2}^{2}$
Now, $\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g (t_{1} + t_{2})$ or $h = \frac{1}{2} g t_{1} t_{2}$
Also, $h = \frac{1}{2} g t^{2} ∴ t^{2} = t_{1} t_{2}$ or $t = \sqrt{t_{1} t_{2}}$ .

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