A truck starts from rest and accelerates uniformly at `2.0 ms^(-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 th (a) velcity , and (b) acceleration of the stone at t = 11 s ? (Neglect air resistance ).

Explanation

If the ball hits truck, for the horizontal motion of truck and ball,
Displacement of ball in horizontal = Displacement of truck in horizontal direction
`v cos 30^@ cc (t - 2) = (1)/(2) at^2`
For the vertical motion of ball, the vertical displacement should be zero.
`v sin 30^@ xx (t - 2) - (1)/(2) g (t - 2)^2 = 0`
`rArr (t - 2) = (2 v sin 30)/(g) = (v)/(g) rArr = 2 + (v)/(g)`
`(sqrt(3 v)^2)/(2) xx (v)/(g) = (1)/(2) a (2 + (v)/(g))^2`
`(sqrt(3 v)^2)/(g) = a(2 + (v)/(g))^2 rArr a = (sqrt(3 v)^2)/(g(2 + (v)/(g))^2)`.