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January 30, 2015

January 30, 2015

Posted by **NeedHelp** on Saturday, February 12, 2011 at 11:57pm.

After how much timeis the shell at a height of 6.20 X 102 and heading down. What equation would be used to solve this?

- physics -
**Henry**, Monday, February 14, 2011 at 1:49amVf^2 = Vo^2 + 2(-9.8)d = 0,

(225)^2 - 19.6d = 0,

50625 - 19.6d = 0,

-19.6d = -50625,

d(up) = 2583m, max.

d = Vot + 0.5gt^2 = 2583 - 620 = 1963m

225t + 4.9t^2 = 1963,

4.9t^2 + 225t -1963 = 0,

Solve using Quadratic Formula and get:

t = 7.49958s, and t = - 48.526.

Use positive value of t.

t(down) = 7.49958s = time to go from

2583m to 620m.

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