Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

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Unformatted text preview: sanchez (es25396) Homework #5 - Projectiles, Net Force, Newtons 2nd Law guevara (130101) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A basketball player 2 . 02 m tall wants to make a basket from a distance of 10 . 1 m. The hoop is at a height of 3 . 05 m. The acceleration of gravity is 9 . 8 m / s 2 . Neglect air friction. 2 . 02 m 3 . 05 m 10 . 1 m v 24 . 7 If he shoots the ball (from a height of 2 . 02 m) at a 24 . 7 angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the back- board? Correct answer: 12 . 9422 m / s. Explanation: Given : h i = 2 . 02 m , h f = 3 . 05 m , and x = 10 . 1 m . The components of the initial velocity are v ix = v x = v i cos and v iy = v i sin . Since x = v x t , the time for the ball to move x horizontally is t = x v x = x v i cos . At this time, the vertical displacement of the ball must be y = v iy t- 1 2 g ( t ) 2 = v i sin x v i cos - g ( x ) 2 2 v i 2 cos 2 = x tan - g ( x ) 2 2 v i 2 cos 2 Thus y- x tan =- g ( x ) 2 2 v i 2 cos 2 2 v i 2 cos 2 = g ( x ) 2 x tan - y v i 2 = g ( x ) 2 2 cos 2 ( x tan - y ) . Since x tan - y = (10 . 1 m) tan 24 . 7 - (1 . 03 m) = (4 . 64548 m)- (1 . 03 m) = 3 . 61548 m , we obtain v i = radicalBigg g ( x ) 2 2 cos 2 ( x tan - y ) = radicalBigg (9 . 8 m / s 2 ) (10 . 1 m) 2 2 (cos 24 . 7 ) 2 (3 . 61548 m) = 12 . 9422 m / s 002 10.0 points A ball is thrown straight upward and returns to the throwers hand after 3 s in the air. A second ball is thrown at an angle of 31 with the horizontal....
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Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

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