A ball is thrown vertically upwards from the ground and a students 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 3s after it was projected from the gorund is [Take `g=10ms^(-2)`]

Options

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

Answer

D

Explanation

Total height `h=15+(u^(2))/(2g)=15+((10)^(2))/(2xx10)` or `h=20m`
Initial velocity `u=sqrt(gh)=sqrt(2xx10xx20)=20m//s`
Now applyint `v=n+at`, have
`v=(+20)+(-10)(3)=-10m//s" ":.` Velocity is 10 m/s downwards