Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc

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 20 m/s at an angle 52 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. 1)What is the horizontal component of the ball’s velocity right before Sarah catches it?2)What is the vertical component of the ball’s velocity right before Sarah catches it? 3)What is the time the ball is in the air? 4)What is the distance between the two girls? 5) 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 16 m above the ground, and takes 2.826 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand? 6) How high above the ground will the ball be when it gets to Julie?

Explanation / Answer

Given,

h = 1.5 m ; vo = 20 m/s ; theta = 52 deg ;

1) Vox = vo cos(theta) = 20 x cos52 = 12.31 m/s

2) Voy = vy sin(theta) = 20 x sin52 = 15.76 m/s

3)time taken to reach the max height be t

t = Voy/g = 15.76/9.8 = 1.61 sec. (using first eq of motion)

total time = T = 2t = 2 x 1.61 = 3.22 s

hence, T = 3.22 s

4)distance between girls = D = Vox x T = 12.31 x 3.22 = 39.64 m

Hence, D = 39.64 m

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