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height of helicopter above ground is given as h = 3t^3, where h is in meters and t is in seconds

height of helicopter above ground is given as h = 3t^3, where h is in meters and t is in seconds. After 2.00s the helicopter drops a bag. How long after is is released does the bag reach the ground?
I know the answer is 7.96s, but do not know how to find it.
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Answer #1
If the height of the helicopter equals the equation given thenafter 2 seconds the helicopter will be h=3*2^3 = 24 metershigh. Using the equation y = y(start) + .5 * 9.81*t   (this should be in your book) We plug in y (which is the distance traveled to   which would be 0 andy(start) which would be 24 meters, then solve for t ending up with4.89297 seconds.   It is wise to note that if there wasan initial velocity then the equation listed would be y =y(start) + v(start)* t + .5 * 9.81 * t
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