Question 5.11:

A truck starts from rest and accelerates uniformly at 2.0 m s -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
(a) velocity, and
(b) acceleration of the stone at t = 11 s? (Neglect air resistance.)


Initial velocity, u = 0, a =0, t = 10 s
Let v be the velocity of the truck when the stone is dropped from it after t = 10 s.
Using the relation, v = u + at, we get
υ = 0 + 2.0 x 10 = 20 m/s
(a) Horizontal velocity of the stone when it is dropped from the truck is
υx = v = 20 m/s
As air resistance is neglected, so υx = constant.
Motion in the vertical direction :
Initial velocity of the stone, υy = 0 at t = 10 s
acceleration, ay = g = 10 m s-2, time t = 11-10 = 1 s
If vy be velocity of the stone after 1 s of drop (i.e. at t = 11 s,) then
vy = uy + ayt = 0 + 10 x 1 = 10 m s-1
If v be the velocity of the stone after 11s, then
 
v=vx2+vy2
=202+102
=500=22.4 ms-1
Let θ be the angle which resultant velocity makes the horizontal direction
Then tanθ=vyvx=1020=0.5
θ=26.60
(b) At the moment, the stone is dropped from the truck, the horizontal force on the stone is zero,so,
ax = 0 and ay = acceleration along vertical direction = +g = 10 m s-2 which acts in downward direction.
If a = resultant acceleration of the stone, then
a=ax2+ay2=02+102
a=10 ms-2    Vertically downward