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.

Given:
Velocity of the truck = 14.7 m/s



Distance covered by the truck when the ball returns again to the truck = 58.8 m



(a) Therefore, we can say that the time taken by the truck to cover 58.8 m distance is equal to the time of the flight of the truck.



Time in which the truck has moved the distance of 58.8 m:



We consider the motion of the ball going upwards.
T = 4 s



Time taken to reach the maximum height when the final velocity v = 0:



v = uat
0 = u + 9.8 × 2
u = 19.6 m/s
19.6 m/s is the initial velocity with which the ball is thrown upwards.


(b) From the road, the motion of ball seems to be a projectile motion.
Total time of flight (T) = 4 seconds



Horizontal range covered by the ball in this time, R = 58.8 m
We know:
Here, α is the angle of projection.


Now,
u cosα = 14.7 ------- (i)



Now, take the vertical component of velocity.



Using the equation of motion, we get:



Here, v is the final velocity.



Thus, we get:



Vertical displacement of the ball:





---------- (ii)


Dividing (ii) by (i), we get:



tan α=1.333




From (i), we get:




Therefore, when seen from the road, the speed of the ball is 25 m/s and the angle of projection is 53° with horizontal.


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