Initial velocity,u=0,a=2.0m s–2,t=10 s
Let v be the velocity of truck when the stone is dropped from it after t = 10 s.
∴ Using the relation, v = u + at, we get
v=0+2.0×10=20 m s−1Horizontal velocity of the stone when it is dropped from the truck is vx=v=20ms–1.
As air resistance is neglected, so vx = constant.
Motion in the vertical direction :
Initial velocity of the stone, vy=0 at t=10s
acceleration, ay=g=10ms–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+10×1=10 m s−1
If v be the velocity of the stone after 11 s, then
v=√vx2+vy2
=√(20)2+(10)2
=√500=22.4 m s−1
Let q be the angle which the resultant velocity OC of the stone makes with the horizontal direction OA i.e. with vx. Then from ΔOAC,
tan‌θ=

AC

OA

=

vy

vx

=

10

20

=0.5
∴θ=26.6°
(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=10ms–2 which acts in downward direction.
∴ If a = resultant acceleration of the stone, then
a=√ax2+ay2
=√02+(10)2 or a=10 ms−2
and it acts vertically downward.