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(B) For a body falling freely from rest $(u = 0)$,the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$.$1$. Distance covered in the firs

Vedclass · Accepted Answer

$h_1 = \frac{h_2}{3} = \frac{h_3}{5}$

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(B) For a body falling freely from rest $(u = 0)$,the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$.$1$. Distance covered in the first $5 \ s$ $(t = 5 \ s)$:$h_1 = \frac{1}{2}g(5)^2 = \frac{25}{2}g$ ... $(i)$$2$. Distance covered in the first $10 \ s$ $(t = 10 \ s)$:$h_1 + h_2 = \frac{1}{2}g(10)^2 = \frac{100}{2}g$ ... (ii)$3$. Distance covered in the first $15 \ s$ $(t = 15 \ s)$:$h_1 + h_2 + h_3 = \frac{1}{2}g(15)^2 = \frac{225}{2}g$ ... (iii)Subtracting $(i)$ from (ii):$h_2 = (h_1 + h_2) - h_1 = \frac{100}{2}g - \frac{25}{2}g = \frac{75}{2}g = 3 \left( \frac{25}{2}g \right) = 3h_1$Subtracting (ii) from (iii):$h_3 = (h_1 + h_2 + h_3) - (h_1 + h_2) = \frac{225}{2}g - \frac{100}{2}g = \frac{125}{2}g = 5 \left( \frac{25}{2}g \right) = 5h_1$Thus,$h_1 = \frac{h_2}{3} = \frac{h_3}{5}$.

$A$ stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first $5 \ s$,the next $5 \ s$ and the next $5 \ s$ respectively. The relation between $h_1, h_2$ and $h_3$ is:

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