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Unformatted text preview: above its launch point has the speed of the ball decreased to onehalf of its initial value? Solution: Given: y max = 16 m Infer: v y,max = 0, y i = 0 Find: y A when v yA = v yi /2 Also, need v yi y y i, v yi y max , v y,max y A, v yA To maximum height (drop y subscript in v): v 2 max = v 2 i –2g(y maxy i ) Solve for v i v 2 i = v 2 max +2g(y maxy i ) = 2gy max To intermediate point: v 2 A = v 2 i –2g(y Ay i ) Solve for y A y A = (v 2 i – v 2 A )/(2g) = [v 2 i – (v i /2) 2 ]/(2g) = v i 2 (11/4)/(2g) = v i 2 (3/4)/(2g) = 3v i 2 /(8g) = 3(2gy max )/(8g) = 3y max /4 = 3(16m)/4 = 12 m Example Problem 4.43 (Hard) On a hot summer day, a young girl swings on a rope above a local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0 o above the horizontal. If she is in flight for 0.616 s, how high above the water was she when she let go of the rope?...
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This note was uploaded on 12/10/2011 for the course PHY 1111 taught by Professor Stencil during the Fall '11 term at UGA.
 Fall '11
 Stencil
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