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 49 degrees;   with respect to the horizontal.  Sarah catches the ball 1.5 meters above the ground.

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.

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

Please show all the steps so I can get it

Explanation / Answer

R = V^2 sin (2 theta) / g    range formula to get separation

R = 12^2 sin 98 / 9.8 = 14.6 m

Can use range formula because final height = initial height

Sarah throws ball back with Vh = 10 m/s    horizontal speed

Ball reaches height of 5.5 m above release point

Vf^2 - Vi^2 = 2 g h     where Vi and Vh are vertical speeds

Vi = (2 g h)^1/2 = (2 * 9.8 * 5.5)^1/2 = 10.4 m/s

Since we know separation and horizontal speed

t = 14.6 / 10 = 1.46 sec    time for ball to reach Julie

h = Vi t - 1/2 g t^2

h = 10.4 * 1.46 - 1/2 * 9.8 * 1.46^2 = 4.74 m

Final height of ball above release point or 6.26 m above ground

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