Question -
Answer -
Answer: (a) 22.36 m/s, at anangle of 26.57° with the motion of the truck
(b) 10 m/s2
(a) Initial velocity of the truck, u =0
Acceleration, a = 2 m/s2
Time, t = 10 s
Asper the first equation of motion, final velocity is given as:
v = u + at
=0 + 2 × 10 = 20 m/s
Thefinal velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontalcomponent (vx) of velocity, in the absence of air resistance, remainsunchanged, i.e.,
vx = 20 m/s
The vertical component (vy) ofvelocity of the stone is given by the first equation of motion as:
vy = u + ayδt
Where, δt = 11 – 10 = 1 s and ay =g = 10 m/s2
∴vy = 0 + 10 × 1 = 10m/s
The resultant velocity (v) of thestone is given as:
Let θ be the angle made bythe resultant velocity with the horizontal component of velocity, vx
=26.57°
(b) When the stone is droppedfrom the truck, the horizontal force acting on it becomes zero. However, thestone continues to move under the influence of gravity. Hence, the accelerationof the stone is 10 m/s2 and it acts vertically downward.