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 travelled 58.8 m . Determine the speed and the angle of projection as seen from the road.

Step-by-step solution

The velocity of the truck is v_t = 14.7 m/s . The time taken by the truck to travel 58.8 m is the time of flight of the ball. T = 58.8 14.7 = 4 s Since the ball returns to the truck, its horizontal velocity as seen from the road must be equal to the velocity of the truck. u_x = v_t = 14.7 m/s The time of flight is given by T = 2u_y g . 4 = 2u_y 9.8 u_y = 19.6 m/s The speed of projection as seen from the road is u = u_x^2 + u_y^2 . u = (14.7)^2 + (19.6)^2 = 216.09 + 384.16 = 600.25 = 24.5 m/s The angle of projection