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 12 m/s at an angle 48 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

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 10 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?

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

5)
Vertical distance covered = 7m - 1.5 m = 5.5 m

At Highest point,
Vertical Velocity = 0
Horizontal velocity = 10 m/s

Let the initial Velocity
Using Newton law of motion for vertical motion-
v^2 = u^2 - 2*a*s
0 = u^2 - 2*9.8*5.5
u = 10.38 m/s
This is the initial Vertical speed of the ball.

Speed of the ball when it leaves Sarah's hand, v = sqrt(10^2 + 10.38^2)
Speed of the ball when it leaves Sarah's hand, v = 14.41

6)
Distance between the two girls, S = 14.61 m
Time taken to reach Julie, = S/Vx
t = 14.61/10 = 1.46 s

S = u*t - 1/2*at^2
S = 10.38*1.46 - 1/2 * 9.8 * 1.46^2
S = 4.71 m

High above the ground will the ball be when it gets to Julie = 1.5 + 4.71
High above the ground will the ball be when it gets to Julie = 6.21 m

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