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Throwing and Catching at Same Height

A stone is th...

Question

A stone is thrown vertically up from the top of a tower with some initial velocity and it arrives on the ground after $t_{1}$ seconds. Now if the same stone is thrown vertically down from the top of the same tower with the same initial velocity, it arrives on ground after $t_{2}$ seconds. How much time will the stone take to reach the ground if it is dropped from the same tower ?

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

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t_{1} + t_{2}

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(t_{1} + t_{2}) / 2

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None of these

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Solution

The correct option is D $\sqrt{t_{1} t_{2}}$
Distance traveled $= h$
$- h = u t_{1} + \frac{1}{2} g t_{1}^{2} ⟶ 1$
$h = u t_{2} + \frac{1}{2} g t_{2}^{2} ⟶ 2$
Where is positive number
Multiplying $1$ by $t_{2}$
and $2$ by $t_{1}$ and adding both equations
$h (t_{1} + t_{2}) = \frac{1}{2} g t_{1} t_{2} (t_{1} + t_{2}) ⟶ 3$
$y = \frac{1}{2} g t^{2} ⟶ 4$
From $3$ and $4$ we get
$t = \sqrt{t_{1} t_{2}}$

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