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put this solution on YOUR website!A helicopter is taking off vertically when a crate slips out of the cargo door. At that instant, the helicopter is moving upward at 15 m/s and is 50 m above the ground. What is the maximum height the crate will reach above the ground?
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Using 1/2 the gravitational acceleration of a falling object; 9.8m/s^2:
h = -4.9x^2 + 15x + 50, where x = time in seconds
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Find the axis of symmetry using x = -b/(2a); in this problem a=-4.9, b=+15
x = -15/(2*-4.9)
x = -15/-9.8
x = +1.53 seconds
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Max height would be the vertex, find that, substitute 1.53 for x:
h = -4.9(1.53^2) + 15(1.53) + 50
h = -4.9(2.34) + 22.95 + 50
h = -11.47 + 22.95 + 50
h = 61.48 meters max height of the wayward box
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After falling from the helicopter, how long will the crate be in the air?
That will be when h = 0
-4.8x^2 + 15x + 50 = 0
Use the quadratic formula; a=-4.9, b=15, c=50
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Do the math here you should get a positive solution of
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x = 5.1 seconds for it to hit the ground
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A graph of this:
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