A stone falls freely under gravity. It covered 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 = 2h_2 = 3h_3`B. `h_1 = (h_2)/(3) = (h_3)/(5)`C. `h_2 = 3h_1` and `h_3 = 3h_2`D. `h_1= h_2 = h_3`

A stone falls freely under gravity. It covered distances `h_1, h_2` an

1.	A stone falls freely under gravity. It covered 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 = 2h_2 = 3h_3`B. `h_1 = (h_2)/(3) = (h_3)/(5)`C. `h_2 = 3h_1` and `h_3 = 3h_2`D. `h_1= h_2 = h_3`
Answer» Correct Answer - B For a particle released from a certain height the distance covered by the particle in relation with time is given by, `h = (1)/(2) g t^2` For first `5 sec, h_1 = (1)/(2) g(5)^2 = 125` Further next `5 sec, h_1 + h_2 = (1)/(2) g (10)^2 = 500` `rArr h_2 = 375` `h_1 + h_2 + h_3 = (1)/(2) g(15)^2 = 1125` `rArr h_3 = 625` `h_1 = 3h_1, h_3 = 5h_1` or `h_1 = (h_2)/(3) = (h_3)/(5)`.

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