Question

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`

Answer 1

Correct Answer - 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 `