50. A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively: The relation between h1, h2 and h3 is

50. 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

\( (a)\; h_1 =\)\( \frac{h_2}{3} \)\( = \)\( \frac{h_3}{5} \)\( \)

\( (b)\; h_2 = 3h_1 \text{ and } h_3 = 3h_2 \)

\( (c)\; h_1 = h_2 = h_3 \)

\( (d)\; h_1 = 2h_2 = 3h_3 \)

Answer :  \( (a)\; h_1 =\)\( \frac{h_2}{3} \)\( = \)\( \frac{h_3}{5} \)\( \)

Total distance in time t is

\( s = \)\(\frac{1}{2}\)\(gt^2 \)

First 5 s:

\( h_1 = \)\(\frac{1}{2}\)\(g(5)^2 = \)\(\frac{1}{2}\)\(g \cdot 25 \)

Next 5 s (5 to 10 s):

\( h_2 = s(10) - s(5) = \)\(\frac{1}{2}\)\(g(100 - 25) = \)\(\frac{1}{2}\)\(g \cdot 75 \)

Next 5 s (10 to 15 s):

\( h_2 = s(15) - s(10) = \)\(\frac{1}{2}\)\(g(225 - 100) = \)\(\frac{1}{2}\)\(g \cdot 125 \)

Therefore,

\( h_1 : h_2 : h_3\) = \(\frac{1}{2}\)\(g \cdot 25 \) : \(\frac{1}{2}\)\(g \cdot 75 \) : \(\frac{1}{2}\)\(g \cdot 125 \) = 25 : 75 : 125 = 1 : 3 : 5

\(\frac{h_1}{h_2}\)\( = \)\(\frac{1}{3} \) \(\rightarrow   h_1 = \)\(\frac{h_2}{3} \)

\(\frac{h_1}{h_3}\)\( = \)\(\frac{1}{5} \) \(\rightarrow   h_1 = \)\(\frac{h_3}{5} \)

\(\rightarrow \: h_1\) = \(\frac{h_2}{3} \) = \(\frac{h_3}{5} \)