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Help me please with this physics question..???I really thankful if anyone of u willing to help me..plzzzz!!!?

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The height of a helicopter above the ground is given by h=3.00t^3,where h in meter and t in second,after 2 s,the helly releases a small mailbag,how long after its realease does the mailbag reach the ground>?...please...i really need it.........

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  1. Position and velocity of the helicopter 2 seconds after take-off:

    h = 3t^3  ------> v = 9t^2

    t = 2 seconds,  h = 3.(2^3)  = 24 meters   and v = 9.(2^2) = 36 meters/second

    The parcel is released at initial height hi = 24 m, with initial velocity vi = 36 m/s upward. .

    Under the action of gravity its movement is described by equation

    h(t) = -(1/2)g.t^2  + vi t + hi

    h(t) = -4.9t^2  + 36 t + 24

    The time it reaches the ground (h = 0) is the positive root of:

    0 = -4.9 t^2 + 36 t + 24 ------> t = 7.3 seconds

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