Given,Â
Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s2
Time, t = 10 s
a)
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/sÂ
The vertical component (vy) of velocity of the stone is given by the first equation of motion as,
           vy = u + ay δtÂ
where,
δt = 11 – 10 = 1 s, andÂ
ay = g = 10 m/s2Â
∴      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,
Therefore,Â
Â
b)Â
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/s2Â and it acts vertically downward.Â
A man of mass 70 kg stands on a weighing scale in a lift which is moving,Â
(a) upwards with a uniform speed of 10 m s–1,Â
(b) downwards with a uniform acceleration of 5 m s–2,Â
(c) upwards with a uniform acceleration of 5 m s–2.
Â
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?