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 stone falls freely under gravity. It covers distances h_(1), h_(2) a

1.	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
Answer» `h_(1) =h_(2) =h_(3)` `h_(1) = 2h_(2) = 3h_(3)` `h_(1) = (h_(2))/(3) - (h_(3))/(5)` `h _(2) =3h_(1) and h _(3) =3H _(2)` Solution :From `d = v_(0) t + 1/2 g t ^(2)` `h = 1/2 g t^(2) (because v _(0) =0)` Now, `h _(1) = 1/2 g (5) ^(2)` `=1/2 (25G)""…(1)` `h _(1) + h_(2) =1/2 g(10) ^(2)` `=1/2 (100g)` `therefore h _(2) = 1/2 (100 g ) - 1/2 (25g)` `=1/2 (75g) =3 [1/2 (25g)]` `= 3h _(1) ""...(2)` `h _(1) + h _(2) + h_(3) = 1/2 g (15) ^(2)` `= 1/2 g (225)` `therefore h _(3) = 1/2 (225g) -1/2 (75g) -1/2 (25g)` `=1/2 (125g)` `= 5 [1/2 (25g)]` `= 5 h_(3) ""...(3)`