A truck starts from rest and accelerates uniformly at `2m*s^(-2)`. At t=10s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground ). What are the
(i) velocity at t=11s?

Explanation

The velocity of the truck after
10s =at =`2xx10=20 *s^(-1)`
So, at the time of release, the stone has a horizontal velocity of `20 m*s^(-1)` , bu no vertical velocity. Its horizontal acceleration =0 and vertical acceleration , `g=9.8 m*s^(-2)`.
(i) At 11 s , i.e., 1 s after release, the horizontal velocity of the stone `=20m*s^(-1)`,
vertical velocity =gt =`9.8 m*s^(-1)`
So, the resultant velocity
`=sqrt((20)^2+(9.8)^2)=22.27 m *s ^(-1)`.