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 21.0 m/s at an angle 58.0 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. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 21.0 m above the ground, and takes 2.886 s to get directly over Julie's head. 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? thanks

Explanation / Answer

The first thing you need to figure out is the time left for the ball to travel to Julie after the ball reaches max height. 1.5m = 16m + (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 the horizontal [x] velocity doesn't change, you can use the velocity given when the ball is at the max height (since at max height the vertical [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 vy = 0 + 9.81 * 0.3153 = 3.092 m/s Then use that to calculate the height of a 1-dimensional free-falling body from 16m for the time that has elapsed post-max height.

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