Question

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

Answer 1

The distance between sarah and julie is not given in the question. You might have calculated in this part in the previously.

Now Vox = 15 m/s

The maximum height possible for sarah throw is y = 15m - 1.5m = 13.5m

The initial speed of the ball leaving sarah would be Voy= Sqrt [ 2 g (13.5) ] = sqrt [ 2 (9.8)(13.5) ] = 16.27m/s

The speed of the ball when it leaves sarah hand is

V= Sqrt [ Vox^2 + Voy^2 ] = sqrt [ 15^2 +16.27^2 ] = 22.13m/s

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Now let d be the distance (not given ) between julie and sarah, so the time taken for ball to reach julie from sarah is

t = d / Vox = d / 15 m/s

now we should use the vertical height equation to identify the ball height when it reach julie (can be higher than julie height).

Y-= Voy. t + (1/2) g t^2

Y = (16.27) ( d / 15 ) + (1/2) (-9.8) ( d/15)^2

once you put the value of d you will know the answer for height of the ball above julie .