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(A) Let the first stone fall from the top of the tower. The distance covered by it in time $t$ is $S_1 = \frac{1}{2} g t^2$.Let the second stone be pr

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$4$

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(A) Let the first stone fall from the top of the tower. The distance covered by it in time $t$ is $S_1 = \frac{1}{2} g t^2$.Let the second stone be projected upwards from the ground. The distance covered by it in time $t$ is $S_2 = u t - \frac{1}{2} g t^2$,where $u = 25 \ m/s$.Since the total height of the tower is $100 \ m$,when the stones meet,the sum of the distances covered by them must be equal to the height of the tower:$S_1 + S_2 = 100$Substituting the expressions for $S_1$ and $S_2$:$\frac{1}{2} g t^2 + (u t - \frac{1}{2} g t^2) = 100$$u t = 100$$25 t = 100$$t = \frac{100}{25} = 4 \ s$.Thus,the two stones will meet after $4 \ s$.

$A$ stone is allowed to fall from the top of a tower $100 \ m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \ m/s$. The two stones will meet after ......... $s$.

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