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 16 m/s at an angle 28 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 16 m/s when it reaches a maximum height of 7 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

Here ,

initial speed , u = 16 m/s

theta = 28 degree

1) horizontal component of the ball's velocity = u * cos(theta)

horizontal component of the ball's velocity = 16 * cos(28)

horizontal component of the ball's velocity = 14.13 m/s

2)

vertical compoent of ball's velocity = u * sin(theta)

vertical compoent of ball's velocity = 16 * sin(28)

vertical compoent of ball's velocity = 7.51 m/s

3)

maximum height of the ball = (u * sin(theta))^2/(2 *g)

maximum height of the ball = (16 * sin(28))^2/(2 * 9.8)

maximum height of the ball = 10.2 m

4)

distance between the girls = u^2 * sin(2 *theta)/g

distance between the girls = 16^2 * sin(2 * 28)/9.8

distance between the girls = 21.7 m

the distance between the girls is 21.7 m

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