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April 18, 2015

April 18, 2015

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

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?

- Physics -
**Damon**, Thursday, August 25, 2011 at 6:05amTo 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

- Physics -
**Damon**, Thursday, August 25, 2011 at 6:15amNow

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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