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August 28, 2015

Homework Help: Physics

Posted by Kyle on Thursday, August 25, 2011 at 5:09am.

A ball is thrown straight up in the air and passes a window 0.30s after being released. It takes 1.5s to go from the window to its maximum height and back down to the window. What was the initial velocity of the ball when it was released?

I found where the ball would be after 0.30s:

Y = V_0 (0.30) - 1/2 (9.8) (0.30)^2
Y = (0.3) (V_0) - 0.441

Then I found where the ball would be after 1.5 which I took to the be same place as the top equation because it returns to the window:

Y = (0.30*V_0) (1.5) - 1/2 (9.8) (1.5)^2
Y = (0.45) (V_0) - 11.025

then I set the equations together and worked it out to get:
V_0 = 70.56 m/s

Is this correct?

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