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From the top of a tower, a particle is t...

From the top of a tower, a particle is thrown vertically downwards with a velocity of `10 m//s`. The ratio of the distances, covered by it in the `3rd` and `2nd` seconds of the motion is `("Take" g = 10 m//s^2)`.

A

`5:7`

B

`7:5`

C

`3:6`

D

`6:3`

Text Solution

Verified by Experts

The correct Answer is:
B
`h_(3)=10+1/2xx10(2xx3-1)=35 m`
`h_(2)=10+1/2xx10(2xx2-1)=25 m`
`h_(3):h_(2)=7:5 `
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