A truck starts from rest and accelerates uniformly at `2.0 m s^(-2)`. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.)

Explanation

(a) Velocity of car ( at t = 10 s ) =` 0 + 2 x× 10 = 20 m s^(-1)`
By the First Law, the horizontal component of velocity is` 20 m s^(-1)` throughout. Vertical component of velocity (at t = 11s) ` = 0 + 10 xx 1 = 10 ms^(-1)`
Velocity of stone (at t = 11s) ` = sqrt( 20^2 + 10^2) = sqrt 500 = 22.4 ms^(-1)` at an angle of `tan^(-1) (1//2)` with the horizontal . (b) `10 ms^(-2)` vertically downwards .