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Answers
Answer:
A) Initial velocity of the truck,u=0
Acceleration,a=2m/s^2
Time,t=10s
As per the first equation of motion,final velocity is given as,
v=u+at
=0+2×10=20m/s
The final velocity of the truck and hence,of stone is 20m/s
At t=11s,the horizontal component (vx)of velocity,in absence of air resistance,remains unchanged,i.e.
vx=20m/s
The vertical component(vy)of velocity of the stone is given by the first equation of motion as:
vy=u+ayt
Where,t=11-10=1s and ay=g=10m/s^2
vy=0+10×=10m/s
The resultant velocity(v)of the stone is given as:
v(vx^2+vy^2)^1/2
(20^2+10^2)^1/2
=22.36m/s
Let be the angle made by the resultant velocity with the horizontal component of velocity vx be thetha
Tan thetha=vy/vx
Thetha=tan^-1 (vy/vx)
Tan^-1 (10/20)
26.570
B)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 stone is 10m/s^2 and it acts vertically downward.