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 15 m/s at an angle 64 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 when it leaves Julie's hand? m/s

2) What is the vertical component of the ball’s velocity when it leaves Julie's hand? m/s

3) What is the maximum height the ball goes above the ground? m

4) What is the distance between the two girls? m

5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 8 m/s when it reaches a maximum height of 13 m above the ground. What is the speed of the ball when it leaves Sarah's hand? m/s

6) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

Explanation / Answer

horizontal component of the velocity = v * cos(theta)

horizontal component of the velocity = 15 * cos(64)

horizontal component of the velocity = 6.57 m/s

vertical component of the velocity = v * sin(theta)

vertical component of the velocity = 15 * sin(64)

vertical component of the velocity = 13.481 m/s

maximum height = v^2 * sin^2(theta) / 2g

maximum height = 15^2 * sin^2(64) / (2 * 9.8)

maximum height = 9.273 m

maximum height above the ground = 9.273 + 1.5

maximum height above the ground = 10.773 m

horizontal range = v^2 * sin(2 * theta) / g

horizontal range = 15^2 * sin(2 * 64) / 9.8

horizontal range = 18.09 m

distance between two girls = 18.09 m

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