SOLUTION: Find the time required for an object to fall to the ground from a building that is 1400 ft high

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Question 99132: Find the time required for an object to fall to the ground from a building that is 1400 ft high
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The equation you are looking for is:
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h+=+-%281%2F2%29%28a%29t%5E2+%2B+v%5Bo%5D%2At+%2B+h%5Bo%5D
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in which the letters represent the following:
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h = the height above ground at the time t (in feet)
t = the elapsed time after the object is released (in seconds)
vo = the initial vertical velocity in feet per second that the object has at release
ho = the initial height above ground when the object is released
a = the acceleration in feet per second^2 due to gravity (usually 32 ft/sec^2 is acceptable)
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The initial vertical velocity (vo) of the object is zero because the object is not thrown
upward or downward. It is just released. Therefore, the term v%5Bo%5D%2At is zero and
can be deleted from the equation.
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The initial height of the object h%5Bo%5D is 1400 feet because that is the height above
ground of the object when it is released.
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After an unknown number of seconds passes, the object hits the ground and at that time
its height above ground is zero. Therefore, we can set h equal to zero.
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With these substitutions into the equation (including that a = 32) the equation becomes:
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0+=+-%281%2F2%29%2A%2832%29%2At%5E2+%2B+1400
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Multiply the -%281%2F2%29%2A%2832%29 to get -16. Substitute this into the equation and it becomes:
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0+=+-16%2At%5E2+%2B+1400
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Get rid of the 1400 on the right side by subtracting 1400 from both sides to make the
equation become:
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-1400+=+-16%2At%5E2
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Divide both sides of this equation by -16, the multiplier of the t%5E2 and the equation
reduces to:
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-1400%2F-16+=+87.5+=+t%5E2
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Solve for t by taking the square root of both sides:
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sqrt%2887.5%29+=+t
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Calculator time ... the square root of 87.5 seconds^2 is 9.354143467 seconds
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So the answer to this problem is about 9.4 seconds
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No sense in having too much accuracy. The acceleration due to gravity on Earth is probably
closer to 32.2 feet per second^2 and it varies depending upon where you are on the globe.
But for most physics problems 32 feet per second^2 is close enough.
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Hope this helps you to understand the problem. If you get into calculus a little (1st
semester probably) you'll find out where the equation that we used comes from.
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