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cw6 - patel(kp9583 CW6 Mackie(20211 This print-out should...

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patel (kp9583) – CW6 – Mackie – (20211) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. DEADLINE IS CENTRAL TIME 001 10.0 points Assume: A 78 g basketball is launched at an angle of 53 . 5 and a distance of 13 . 1 m from the basketball goal. The ball is released at the same height (ten feet) as the basketball goal’s height. A basketball player tries to make a long jump-shot as described above. The acceleration of gravity is 9 . 8 m / s 2 . What speed must the player give the ball? Correct answer: 11 . 5865 m / s. Explanation: Basic concepts: Horizontally, v oh = v cos θ v h = v oh d = v oh t Vertically, v ov = v sin θ v v = v ov - g t h = v ov t - 1 2 g t 2 . Solution: At the maximum range of the ball, v fv = - v ov , so, - v ov = v ov - g t - 2 v ov = - g t t = 2 v ov g . The maximum distance covered is d = v oh t = 2 v oh v ov g d = 2 v cos θ v sin θ g d = v 2 (2 sin θ cos θ ) g = v 2 sin(2 θ ) g .

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