A body falls freely from the top of a tower. | vedclass

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(D) Let the total height of the tower be $h$ and the total time taken to reach the ground be $t$ seconds.Since the body falls freely,the initial veloc

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$125$

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(D) Let the total height of the tower be $h$ and the total time taken to reach the ground be $t$ seconds.Since the body falls freely,the initial velocity $u = 0$.Using the equation of motion $h = \frac{1}{2}gt^2$ ... $(i)$.In the last second,the body covers $36\%$ of the total height,which is $0.36h$.Therefore,in the first $(t - 1)$ seconds,the body covers the remaining height: $h - 0.36h = 0.64h$.Using the equation of motion for $(t - 1)$ seconds: $0.64h = \frac{1}{2}g(t - 1)^2$ ... $(ii)$.Dividing equation $(ii)$ by equation $(i)$:$\frac{0.64h}{h} = \frac{\frac{1}{2}g(t - 1)^2}{\frac{1}{2}gt^2}$$0.64 = \frac{(t - 1)^2}{t^2}$Taking the square root on both sides: $0.8 = \frac{t - 1}{t}$.$0.8t = t - 1 \implies 0.2t = 1 \implies t = 5 	ext{ s}$.Substituting $t = 5$ in equation $(i)$ (taking $g = 10 	ext{ m/s}^2$):$h = \frac{1}{2} 	imes 10 	imes (5)^2 = 5 	imes 25 = 125 	ext{ m}$.

$A$ body falls freely from the top of a tower. It covers $36\%$ of the total height in the last second before striking the ground level. The height of the tower is........$m$

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