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 :

Options

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

B

Explanation

Distance covered in first 5 sec.
`h_(1)=0+(1)/(2)a(5)^(2)`
`h_(1)=(25a)/(2)" " `…(1)
distance covered in first 10 sec
`S_(2)=0+(1)/(2)a(10)^(2)=(100a)/(2)`
So distance covered in second 5 sec.
`h_(2)=S_(2)-h_(1)=(100a)/(2)-(25a)/(2)=(75a)/(2) " " ` ...(2)
distance covered in first 15 sec.
`S_(3)=0+(1)/(2)a(15)^(2)=(225a)/(2)`
so distance covered in last 5 sec.
`h_(3)=S_(3)-S_(2)=(225a)/(2)-(100a)/(2)=(125a)/(2) " " ` ...(3)
using (1), (2) and (3) equation.
`(h_(1))/((25a)/(2))=(h_(2))/((75a)/(2))=(h_(3))/((125a)/(2))`
`h_(1)=(h_(2))/(3)=(h_(3))/(5)`