In the previous problem, distance covered in the `7^(th)` second is

Options

A. 1.25 m
B. 2.5 m
C. 3.75 m
D. 5.0 m

Answer

B

Explanation

`v=u-g t`
`0=65-10 t`
`t=6.5s`

`7^(th)` second consists of `1/2 s` in upward and `1//2 s `in downward motion. Displacement in the last `1//2 s`
`h=1/2g (1//2)^(2)=1.25 m`
Distance in `7^(th)` second`=2h=2.5 m`