A ball is thrown vertically upwards from the ground and a student gazing out of the window sees it moving upward past him at `10ms^-1.` The window is at 15 m above the ground level. The velocity of ball 3 s after it was projected from the ground is [Take `g= 10 ms^-2`]

Options

A. `10 m//s,` up
B. `20 m//s,` up
C. `20 ms^-1,`down
D. `10 ms^-1,` down

Answer

D

Explanation

Total height=`15+(u^2)/(2g)`
`=15+((10)^2)/(2xx10)`
or `h=20m`
initial velocity `u=sqrt2gh`
`sqrt(2xx10xx20)`
`=20 m//s`
Now applying `v=u+at,` we have
`v=(+20)+(-10)(3)`
`:.` Velocity is `10m//s,` downwards.