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(D) Let the height of the tower be $h = 60 \ m$. The meeting point is at a height $y = \frac{2h}{3} = \frac{2 	imes 60}{3} = 40 \ m$ from the ground.

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Let the height of the tower be $h = 60 \ m$. The meeting point is at a height $y = \frac{2h}{3} = \frac{2 	imes 60}{3} = 40 \ m$ from the ground.For the stone dropped from the top: The distance covered is $s_1 = h - y = 60 - 40 = 20 \ m$. Using $s = ut + \frac{1}{2}gt^2$ with $u = 0$,we get $20 = 0 + \frac{1}{2}(10)t^2$,which simplifies to $20 = 5t^2$,so $t^2 = 4$ and $t = 2 \ s$.For the stone projected upwards from the ground: The distance covered is $s_2 = y = 40 \ m$. Using $s = ut - \frac{1}{2}gt^2$,we get $40 = u(2) - \frac{1}{2}(10)(2)^2$.$40 = 2u - 20$.$2u = 60$,which gives $u = 30 \ ms^{-1}$.

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