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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
Distance travelled in nth second
`S_n = u + a/2 (2n - 1)`
`S_3/S_2 = (10 + 10/2 (6-1) )/(10 + 10/2 (4-1)) = 35/25 = 7/5 `
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