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(C) Let the height of the tower be $h$ and the acceleration due to gravity be $g$. The time taken to fall freely from height $h$ is given by $h = \fra

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$t = \sqrt{t_1 t_2}$

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(C) Let the height of the tower be $h$ and the acceleration due to gravity be $g$. The time taken to fall freely from height $h$ is given by $h = \frac{1}{2}gt^2$,so $t = \sqrt{\frac{2h}{g}}$.For the stone thrown upwards with velocity $V_0$,the displacement is $-h$:$-h = V_0 t_1 - \frac{1}{2}g t_1^2 \implies h = \frac{1}{2}g t_1^2 - V_0 t_1$. Since $h = \frac{1}{2}gt^2$,we have $\frac{1}{2}gt^2 = \frac{1}{2}g t_1^2 - V_0 t_1 \implies V_0 t_1 = \frac{1}{2}g(t_1^2 - t^2) \dots (1)$.For the stone thrown downwards with velocity $V_0$,the displacement is $-h$:$-h = -V_0 t_2 - \frac{1}{2}g t_2^2 \implies h = V_0 t_2 + \frac{1}{2}g t_2^2$. Since $h = \frac{1}{2}gt^2$,we have $\frac{1}{2}gt^2 = V_0 t_2 + \frac{1}{2}g t_2^2 \implies V_0 t_2 = \frac{1}{2}g(t^2 - t_2^2) \dots (2)$.Dividing equation $(1)$ by $(2)$:$\frac{V_0 t_1}{V_0 t_2} = \frac{\frac{1}{2}g(t_1^2 - t^2)}{\frac{1}{2}g(t^2 - t_2^2)} \implies \frac{t_1}{t_2} = \frac{t_1^2 - t^2}{t^2 - t_2^2}$.Cross-multiplying: $t_1 t^2 - t_1 t_2^2 = t_2 t_1^2 - t_2 t^2$.$t^2(t_1 + t_2) = t_1 t_2(t_1 + t_2)$.$t^2 = t_1 t_2 \implies t = \sqrt{t_1 t_2}$.

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