Three particles are projected upwards with initial speeds `10 m//s,20 m//s,30 m//s`. The displacements covered by them in their last second of motion are `x_(1),x_(2),x_(3)` then:

Options

A. `x_(1):x_(2):x_(3)=1:2:3`
B. `x_(1):x_(2):x_(3)=1:4:9`
C. `x_(1):x_(2):x_(3)=1:5:7`
D. none of these

Answer

D

Explanation

Distance travelled by each particles in last second of motion i.e. Downwards is equal to the distance travelled by it in first second of its motion i.e. upwards
So, `x_(1)=10-1//2 xx10xx1^(2) =5 m`
`x_(2)=20 -1//2xx10xx1^(2)= 15 m`
`x_(3)=30 -1//2 xx10xx1^(2) =25 m`
So, `x_(1):x_(2):x_(3)=1:3:5`