Question 180633
Equation of height of object (earth):
Because Earth gravitational pull is 32 feet/sec^2
{{{y = -16t^2 + 27t + 6}}}
where
y is the position of the object
t is time in seconds
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Setting y to 0, (which is equivalent to asking when is is level w/ground)
{{{0 = -16t^2 + 27t + 6}}}
Applying the quadratic equation we get:
y = {-0.199, 1.886}
Throw out the negative solution we get
1.886 secs in the air (Earth)
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Similarly, for the Moon (where gravity is 1/6th that of the Earth) we have:
{{{y = -(16/6)t^2 + 27t + 6}}}
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Setting y to 0, (which is equivalent to asking when is is level w/ground)
{{{y = -(16/6)t^2 + 27t + 6}}}
Applying the quadratic equation we get:
y = {-0.218, 10.343}
Throw out the negative solution we get
10.343 secs in the air (Moon)
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CONCLUSION: IT WOULD TAKE LONGER FOR THE OBJECT TO HIT THE GROUND ON THE MOON.
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Below are the details of BOTH quadratic:
Earth's quadratic:
*[invoke quadratic "t", -16, 27, 6 ]
Moon's quadratic:
*[invoke quadratic "t", -2.6666666666666666666666666666667, 27, 6 ]
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