Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1531046 • Letter: J
Question
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 19 m/s at an angle 46 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
B) What is the vertical component of the ball’s velocity right before Sarah catches it?
C) What is the time the ball is in the air?
D) After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 13 m above the ground, and takes 2.452 s to get directly over Julie's head.
What is the speed of the ball when it leaves Sarah's hand?
E) How high above the ground will the ball be when it gets to Julie?
Explanation / Answer
Vox = Vo(cos ) = 19(cos 46) 13.20 m/s ANS-1
Voy = Vo(sin ) = 19(sin 46°) 13.67 m/s ANS-2
time to reach max height = t = Voy/g = 13.67/9.81 = 1.393 s
max height above release point = 1/2gt² = (4.905)(1.393)² = 9.517 m
max height above ground = 9.517 + 1.5 11.02
total flight time of ball = 2t = (2)(1.393) = 2.786 s
distance between two girls = (2.786)(Vox) = (2.786)(13.20) 36.775 m ANS
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