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.

Sol. . a) As seen from the truck the ball moves vertically upward comes back. Time taken = time taken by truck to cover 58.8 m. ∴ time = s/v = 58.8/14.7 = 4 sec. (V = 14.7 m/s of truck) u = ?, v = 0, g = –9.8 m/s2 (going upward), t = 4/2 = 2 sec. v = u + at ⇒ 0 = u – 9.8 × 2 ⇒ u = 19.6 m/s. (vertical upward velocity). b) From road it seems to be projectile motion. Total time of flight = 4 sec In this time horizontal range covered 58.8 m = x ∴ X = u cos θ t ⇒ u cos θ = 14.7 …(1) Taking vertical component of velocity into consideration. y = (0^2- (19.6)^2)/(2 x (-9.8)) = 19.6 m [from (a)] ∴ y = u sin θ t – 1/2 gt2 ⇒ 19.6 = u sin θ (2) – 1/2 (9.8)22 ⇒ 2u sin θ = 19.6 × 2 ⇒ u sin θ = 19.6 …(ii) (u sin⁡θ)/(u cos⁡θ ) = tan θ 19.6/14.7 = 1.333 ⇒ θ = tan–1 (1.333) = 53° Again u cos θ = 14.7 ⇒ u = 14.7/(u cos⁡〖53°〗 ) = 24.42 m/s. The speed of ball is 42.42 m/s at an angle 53° with horizontal as seen from the road.