```Question 115783
First problem:
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You are given the equation of the height of an object that is thrown upward as:
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{{{h(t)=-16t^2+308t}}}
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and the first question you are asked is to find the time that the object reaches its maximum
height. The object starts out at ground level, rises to a peak, and then falls back to
ground level. Neglecting air resistance and other minor considerations, it spends half its
time rising and half its time falling back to ground level. So one way you can find the
time it takes to reach its peak height is to find the time of launch and the total time that
goes by until it hits the ground ... then divide that time by 2.
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Notice that at ground level the height given by h(t) is zero. So let's substitute zero
into the equation for h(t) and get (after reversing the sides of the equation to get it
into a little more familiar form):
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{{{-16t^2 + 308t = 0}}}
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Note that t is a common factor of both the terms on the left side, so it can be factored to
make the left side become:
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{{{t*(-16t+308) = 0}}}
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Notice that this equation will be true if either of the factors is equal to zero because a
multiplication by zero on the left side will make the left side equal the zero on the right
side.
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Setting the first factor [which is t] equal to zero results in:
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{{{t = 0}}}
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This means that at t = 0 seconds the object is at ground launch. No surprise here.
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Next setting the second factor [which is -16t + 308] equal to zero gives:
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{{{-16t + 308 = 0}}}
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Solve this by first subtracting 308 from both sides to get:
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{{{-16t = -308}}}
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and then dividing both sides by -16 to get:
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{{{t = -308/-16 = +19.25}}}
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This tells us that from time of launch at t = 0, 19.25 seconds later the object hits the ground.
Since half of that time was spent rising and half of that time was spent falling back down,
the time at which the object reaches its peak is {{{19.25/2 = 9.625}}}. So at 9.625 seconds
after launch the object is at its maximum height.
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Another way you can find this time is to apply part of the quadratic formula. Recall that
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{{{at^2 + bt + c = 0}}}
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If you compare this with your height equation you will see that a = -16, b = 308, and c = 0.
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Then just use that portion of the quadratic formula that is {{{t = -b/(2*a)}}} to find the
time at the peak. Substituting 308 for b and -16 for a results in:
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{{{t = (-308)/(2*(-16)) = -308/-32 = 9.625}}} seconds.
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This is the same answer, just a little different way of getting it.
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Then to find the maximum height, just substitute 9.625 seconds for t in the height equation
to get:
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{{{h(t) = -16*(9.625)^2 + 308*(9.625) = -16*(92.640625) + 2964.5 = -1482.25 + 2964.5 = 1482.25}}}
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So the object rises to a height of 1482.25 feet. That's quite a throw!!!! Check your problem
to see if the 308 is the correct multiplier of the t term. If it is, that's really the
answer .... 1482.25 feet up and 1482.25 feet back down again.
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Next problem.
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The Area of a square that has S as the length of one side is given by the equation:
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{{{A = S^2}}}
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and the Perimeter of the square is the sum of the lengths of all its sides:
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{{{P = S + S + S + S = 4*S}}}
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Since the Area is 12 more than the Perimeter, if you take 12 away from the Area, the result
will equal the Perimeter. In equation form this is:
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{{{S^2 - 12 = 4S}}}
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Get this into the standard quadratic form by subtracting 4S from both sides to get:
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{{{S^2 - 4S - 12 = 0}}}
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This equation factors to:
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{{{(S +2)(S - 6) = 0}}}
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As in the previous problem, this equation will be true if either factor equals zero.
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Setting the first factor equal to zero results in:
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{{{S + 2 = 0}}}
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Subtract 2 from both sides and you get:
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{{{S = -2}}}
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This answer doesn't make sense ... a side of minus 2 length??? Ignore it.
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Setting the second factor equal to zero gives:
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{{{S - 6 = 0}}}
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Add 6 to both sides and you have:
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{{{S = 6}}}
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This looks better. A square with a side of 6 has an area of 6^2 = 36 and a perimeter of
6*4 = 24. The area is 12 more than the perimeter, just as the problem specified. So the
side length you were to find is S = 6.
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Hope this helps with your understanding of these two problems. It's pretty late so you had
better use your calculator to check my math. The process is correct, but I may have let
a calculation error slip in ... I don't think so but better safe than sorry ...
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