Physics
posted by Kyle .
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?

To check it throw a ball up at 70.56 m/s
a = 9.8
v = 70.56  9.8 t
h = 0 + 70.56 t  4.9 t^2

at the window on the way down, t = .3 + 1.5 up + 1.5 down = 3.3
v = 70.56  9.8(3.3) =  38.22
h = 70.56(3.3)  4.9(3.3)^2 = 179.5

so how long would it take to get to the window on the way up?
179.5 = 70.56 t  4.9 t^2
no way that is t = .3 s 
Now
a = 9.8
v = Vi  9.8 t
h = 0 + Vi t  4.9 t^2


at the top, t = .3 + 1.5/2 = 1.05
and v = 0 = Vi  9.8(1.05)
so
Vi = 9.8(1.05) = 10.29 m/s
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