Question

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection
(a)as seen from the truck,
(b)as seen from the road.

Answer 1

the time taken by the track to move 58.8 m=58.8/14.7=4s
as seen from the truck...
there is no horizontal movement of the ball...so the horizontal component of the ball's velocity is zero...so the angle of projection is 90 degree as seen from truck..
and the velocity be v..
s=ut-1/2gt^2
here s=0
t=4
u=1/2gt=1/2*10*4=20 m/s

as seen from road...
there is a horizontal velocity of the ball...and that is the same that of the truck...=14.7 m/s
and a vertical velocity of 20 m/s

Answer 2

Thank you.

Answer 3

welcome:)