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A stone falls freely under gravity. It c...

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

1) `h_(1)=2h_(2)=3h_(3)`

B

2) `h_(1)=frac(h_(2))(3)=frac(h_(3))(5)`

C

3) `h_(2)=3h_(1)andh_(3)=3h_(2)`

D

4) `h_(1)=h_(2)=h_(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
For a particle released from a certain height the dia=stance covered by the particle in relation with time is given by
`h=(1)/(2)gt^(2)`
For first 5 sesc, `h_(1)=(1)/(2)g(5)^(2)=125`
Further next 5sec
`H_(1)+(h_(2)(1)/(g)(10)^(2)=500`
`Rightarrowh_(2)=375`
`H_(1)+h_(2)+h_(3)=(1)/(2)g(15)^(2)=1125`
`Rightarrowh_(3)=625`
`h_(1)=3h_(1)h_(3)=5h_(1)` or `h_(1)= (h_(2)/(3)= (h_(3)/(5)`
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