Posted by NeedHelp on .
A cannon shell is fired up in the air at an initial speed of 225m/s.
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?
Vf^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.