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 20 m/s at an angle 50 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. 1) 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 15 m/s when it reaches a maximum height of 15 m above the ground. What is the speed of the ball when it leaves Sarah's hand? 2) 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

The first thing you need to figure out is the time left for theball to travel to Julie after the ball reaches max height.

1.5m = 1.5m + (1/2) * (-9.81) * t^2

Which should give you t-up = 1.7194 seconds

Then take the total distance between Sarah and Julie and calculate the total time the ball will take to get there. Since thehorizontal [x] velocity doesn't change, you can use the velocitygiven when the ball is at the max height (since at max height thevertical [y] velocity is zero).

26.451m / 13m/s = 2.0347 seconds

After that, you just need to find the difference between the two times:

2.0347s - 1.7194s = 0.3153 seconds <= Time the ball is in flightafter max height.

And find the velocity at that time:

vy = vy0 + at => vy = 0 + 9.81 * 0.3153 = 3.092 m/s

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