Question

a person standing on truck moving with a constant velocity of 14.7m/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 58.8m .find the speed and the angle of projection (a)as seen from the truck(b)as seen from the road.

Answer 1

we know that

velocity (v) = distance travelled (s) / time taken (t)

so,

time taken by truck to travel the distance of 58.8m at a velocity of 14.7 m/s will be

t = s/v = 58.8 / 14.7

so,

t = 4s

now, the time of flight of the ball projected should be equal to this time 't'.

..

(a)

From the trucks point of view the ball will only have a vertical velocity component.

so, the velocity will be

v = u_y = (1/2)gt

so,

v = (1/2) x 10 x 4

thus,

v = 20 m/s

..

(b)

now,

from the ground's perspective the ball will have both vertical as well as horizontal velocity.

here,

vertical velocity ; u_y = 20 m/s

horizontal velocity ; u_x = 14.7 m/s

so,

the resultant velocity of the ball will be

v = [u_x² + u_y²]^1/2 = [14.7² + 20²]^1/2

= [616.09]^1/2

thus,

v = 24.82 m/s