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Vertical Motion of Objects

  1. Oct 6, 2009 #1
    The height of a helicopter above the ground is given by h = 3.50t^3, where h is in meters and t is in seconds. After 1.85 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? (Ans. in s.)

    (1) y-y0=v0t+1/2*gt^2

    (2) v^2=v0^2+2g(y-y0)

    (3) y-y0=1/2*(v0+v)t

    I sub. 1.85s for t in h=3.50t^3 and h=y0=22.2m.. This is the initial position. The final position is y=0. So I substituted all of my known values into (1) and got v0= -2.92m/s(initial velocity)
    then I sub. all the known values into (2) and got v= -21.1m/s. Finally, I sub. v, v0, and the rest of the known values into (3) and got 1.85s. I believe it is the wrong answer. Please help me to solve this problem.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 6, 2009 #2


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    Homework Helper

    I agree with your initial height.
    But you can't use a constant acceleration formula to find the velocity of the helicopter at time 1.85 s. There are no formulas for height increasing as the cube of time. If you know a bit of calculus, you can differentiate to get a formula for velocity. If not, you'll have to carefully graph y = 3.50t^3 and draw a tangent line at time 1.85 to estimate the slope, which is the velocity.
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