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)`.