A truck starts from rest and accelerates uniformly at 2.0 m . 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 is the magnitude of acceleration (in m ) of the stone at t = 11s? (Neglect air resistance.)
When the stone is dropped from the truck , the horizontal force acting on it become zero after stone is released. Stone continue to move under the influence of gravity only. So that acceleration of the stone is equal to the gravitational acceleration g.
a = g = 10ms-2 acting vertically downward.