Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s2
Time, t = 10 s
According to the equation of motion, we have
v = u + at
= 0 + 2x10
= 20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (vx) of velocity, in the absence of air resistance, remains unchanged.
i.e., vx = 20 m/s
The vertical component (vy) of velocity of the stone is given by the first equation of motion as,
vy = u + ay δt
δt = 11 – 10 = 1 s, and
ay = g = 10 m/s2
∴ vy = 0 + 10 × 1
= 10 m/s
The resultant velocity (v) of the stone is given as:
Let, be the angle made by the resultant velocity with the horizontal component of velocity vx,
When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is 10 m/s2 and it acts vertically downward.