Question

A truck starts from rest and accelerates uniformly with 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 the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance).

Answer 1

(a) As v = u + at, when v = v_x, u = 0

a = 2ms^-2, t = 10 s,

we get v = 0 + 2 x 10 = 20 ms^-1

v_y = 0 + 9.8 x 0.1 = 0.98 ms^-1

[Time after dropping out of the top of the trade = 10.1 - 10 = 0.1 s]

v_R = √{v²_x + v²_y} = √{(20)² + (0.98)²} = 20.02 ms^-1

and tanθ = v_y/v_x = 0.98/20 = 0.049

or, θ = tan^-1 (0.049) = 2.8°

(b) Acceleration at 10.1 s = g = 9.8 ms^-2