Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1326357 • 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 16 m/s at an angle 30 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
What is the vertical component of the ball’s velocity when it leaves Julie's hand?
What is the maximum height the ball goes above the ground?
What is the distance between the two girls?
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 16 m/s when it reaches a maximum height of 8 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
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
Here ,
initial speed , u = 16 m/s
theta = 30 degree
horizontal component of ball's velocity in Julie's hand = 16 * cos(30)
horizontal component of ball's velocity in Julie's hand = 13.9 m/s
the horizontal component of ball's velocity in Julie's hand is 13.9 m/s
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vertical component of ball's velocity in Julie's hand = 16 * sin(30)
vertical component of ball's velocity in Julie's hand = 8 m/s
vertical component of ball's velocity in Julie's hand is 8 m/s
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maximum height above ground = vy^2/(2 *g) + 1.5
maximum height above ground = 8^2/(2 *9.8) + 1.5
maximum height above ground = 4.77 m
the maximum height above ground is 4.77 m
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distance between girls = 2 * Vx * Vy/g
distance between girls = 2 * 8 * 13.9/9.8
distance between girls = 22.7 m
the distance between girls is 22.7 m
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