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Time of Flight

A body is thr...

Question

A body is thrown upward at an angle $θ$ with horizontal from a point A on top of a tower. It reaches ground in time $t_{1}$ . If it is thrown downward at an angle $θ$ with horizontal from same point with same speed, it reaches the ground in time
$t_{2}$ . If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

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

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\frac{(t_{1} + t_{2}]}{2} sin θ

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

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

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Solution

The correct option is C $\sqrt{t_{1} t_{2}}$
If velocity of the body is $u_{0}$ , then vertical component of velocity will be $u_{0} sin θ$ .
Sign convention - Taking upward direction from point A as positive .
When ball is thrown upward $- h = (u_{0} sin θ) t_{1} - \frac{g {t_{1}}_{1}^{2}}{2}$ $\to$ (i)
When ball is thrown downward $- h = - (u_{0} sin θ) t_{2} - \frac{g {t_{2}}_{2}^{2}}{2}$ $\to$ (ii)
Subtracting (ii) from (i) , we get $0 = (u_{0} sin θ) (t_{1} + t_{2}) + \frac{g ({t_{2}}_{2}^{2} - {t_{1}}_{1}^{2})}{2}$ $\Rightarrow u_{0} sin θ = \frac{g (t_{1} - t_{2})}{2}$ .
Putting above in equation (ii), we get $- h = - \frac{g (t_{1} - t_{2}) t_{2}}{2} - \frac{g {t_{2}}_{2}^{2}}{2} \Rightarrow h = \frac{g t_{1} t_{2}}{2}$ .
For free fall initial velocity = 0 $\Rightarrow h = \frac{g t^{2}}{2}$ .
Equating h, we get $\frac{g t_{1} t_{2}}{2} = \frac{g t^{2}}{2} \Rightarrow t = \sqrt{t_{1} t_{2}}$

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