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

Options

A. `h_(1) =h_(2) =h_(3)`
B. `h_(1) = 2h_(2) = 3h_(3)`
C. `h_(1) = (h_(2))/(3) - (h_(3))/(5)`
D. `h _(2) =3h_(1) and h _(3) =3h _(2)`

Answer

C

Explanation

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)`