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Physics 18 Online
OpenStudy (anonymous):

The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 8.0 m/s when it reaches a maximum height of 13.0 m above the ground. What is the speed of the ball when it leaves Sarah's hand?

OpenStudy (anonymous):

the fact that it's moving at its highest point means that the x velocity is 8 m/s (because vertical velocity is zero at that point) you can find Vyi from Vyf^2 = Vyi^2 -2*g*11.5m (where Vfy is going to be zero) calculate initial speed from the two components of initial velocity we just found |Vi| = sqrt ( Vyi^2 +Vix^2)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

no idea.

OpenStudy (anonymous):

ok thank you so much though u really helped me!

OpenStudy (anonymous):

sure:)

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