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A stone falls freely under gravity. It covers distances $h_1$, $h_2$ and $h_3$ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between $h_1$, $h_2$ and $h_3$ is:
A
$h_1 = h_2 = h_3$
B
$h_1 = 2h_2 = 3h_3$
C
$h_1 = \frac{h_2}{3} = \frac{h_3}{5}$
D
$h_2 = 3h_1$ and $h_3 = 3h_2$
Detailed Solution
$h_1 = \frac{1}{2}g(5)^2$, $h_1 + h_2 = \frac{1}{2}g(10)^2$ and $h_1 + h_2 + h_3 = \frac{1}{2}g(15)^2$
So $h_1 : h_2 : h_3 = 1 : 3 : 5$, i.e. $h_1 = \frac{h_2}{3} = \frac{h_3}{5}$
So $h_1 : h_2 : h_3 = 1 : 3 : 5$, i.e. $h_1 = \frac{h_2}{3} = \frac{h_3}{5}$