Question 112941
The equation for the height of a "free falling body in motion" is:
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{{{s = -16t^2 + v[o]t + s[o]}}}
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where:

{{{s}}} is the height of the object at a time equal to or greater than zero;
{{{t}}} is the time at or after release of the object
{{{v[o]}}} is the initial velocity of the object at launch (+ is up and - is down}
{{{s[o]}}} is the height from which the object is launched
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The problem tells you that {{{v[o] = +64}}} and {{{s[o] = 25}}}. Substitute these values and
the height equation becomes:
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{{{s = -16t^2 + 64t + 25}}}
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and this is the answer to part a of your problem.
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part b asks how high the object is after 1 second. Just plug 1 in for t and you get:
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{{{s = -16*(1^2) + (64*1) + 25 = -16 + 64 + 25 = 73}}}
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So after 1 second the object is 73 feet above the ground.
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part c asks at what time the object will reach its maximum height. This can be found by
recognizing that the height equation is a quadratic equation of the standard form:
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{{{s = at^2 - bt + c}}}
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if the quadratic formula is applied to solve this equation, the vertex of the path will occur
where {{{t = -b/(2a)}}} 
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By comparing terms of the equation that is the answer for part a with the terms of the
standard form of the quadratic equation, you find that a = -16 and b = +64. Therefore,
you can get that the vertex occurs at a time equal to:
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{{{t = -b/(2a) = -(+64)/(2*-16) = -64/-32 = 2}}}
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So the peak height occurs at t = 2 seconds after launch of the object. This is the answer
to part c.
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part d asks how high the object gets at its peak. We have found that the answer occurs
when t is 2 seconds. So all that we have to do to get the peak height is to return to
the equation we got for part a and substitute 2 for 2. This is done as follows:
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Start with:
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{{{s = -16t^2 + 64t + 25}}}
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Substitute 2 for t and you get:
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{{{s = -16*(2^2) + 64*2 + 25 = (-16*4) + (64*2) + 25 = -64 + 128 + 25 = 89}}}
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The answer is that the object reaches a maximum height above ground of 89 feet.
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Hope this helps you to understand the problem.
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