physics

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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?

  • physics -

    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.

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