This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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?...
View
Full Document
 Fall '11
 Stencil
 Work, Velocity, 8g, 2.25 m/s, Resultant vector magnitude

Click to edit the document details