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A truck starts from rest and accelerates...

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
(ii) acceleration of the stone at t=11 s ? (Neglect air resistance. )

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Verified by Experts

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)`.
(ii) The acceleration of the stone = downward acceleration due to gravity =`9.8 m*s^(-2)`.
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