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Throwing and Catching at Same Height

A body is thr...

Question

A body is thrown vertically upward from the top $A$ of tower. It reaches the ground in $t_{1} s$ . If it is thrown vertically downwards from $A$ with the same speed it reaches the ground in $t_{2} s$ . If it is allowed to fall freely from $A$ , then the time it takes to reach the ground is given by

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

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

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

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t = \sqrt{\frac{t_{1}}{t_{2}}}

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Solution

The correct option is C $t = \sqrt{t_{1} t_{2}}$
Taking downward direction as positive direction.
Case 1: Ball is thrown upwards with velocity $u$ .
Using second equation of motion we get
$+ h = - u t_{1} + \frac{1}{2} g t_{1}^{2} \dots \dots (i)$
Case 2: Ball is thrown downwards with velocity $u$ .
Using second equation of motion we get
$+ h = u t_{2} + \frac{1}{2} g t_{2}^{2} \dots \dots (i i)$
Multiplying Eq. $(i)$ by $t_{2}$ and Eq. $(i i)$ by $t_{1}$ and adding. We get
$h (t_{1} + t_{2}) = \frac{1}{2} g t_{1} t_{2} (t_{1} + t_{2})$
or $h = \frac{1}{2} g t_{1} t_{2} \dots \dots (i i i)$
In case of free fall, we can write
$h = \frac{1}{2} g t^{2} \dots \dots (i v)$
$∴ t^{2} = t_{1} t_{2}$
$t = \sqrt{t_{1} t_{2}}$

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