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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 cosθ t ⇒ u cosθ = 14.7 …(1) Taking vertical component of velocity into consideration. y = (02-19.62) / (2 x (-9.8)) = 19.6 m [from (a)] ⇒ y = u sinθ t – 1/2 gt2 ⇒ 19.6 = u sinθ (2) – 1/2 (9.8)22 ⇒ 2u sinθ = 19.6 × 2 ⇒ u sinθ = 19.6 …(ii) usinθ / ucosθ = tanθ = 19.6 / 14.7 = 1.333 ⇒ θ = tanθ–1 (1.333) = 53° Again u cosθ = 14.7 ⇒ u = cos53 = 24.42 m/s. The speed of ball is 42.42 m/s at an angle 53° with horizontal as seen from the road.
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