Kenneth S. answered • 01/30/17
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a) which technique for solving the quadratic equation is going to be best for this equation and why?
s(t)= -16t^2 +64t +160 = 0 is the equation to be solved, to find time it takes for projectile to strike ground.
FACTOR: -16(t2 - 4t - 10) = 0; we see that factoring will not give a solution, so use the quadratic formula on the quadratic enclosed in parentheses. Do not accept a negative answer, because t≥0 is an obvious domain restriction.
d) the ball will never reach a height of 300 ft. How can this be determined algebraically?
Find the vertex of the quadratic, which is the function's maximum. t = -b/(2a) = -(-4)/2 = 2, and
s(2)= -16(22 -4•2-10) = -16(4 - 18) = 16(14) = 224, which is maximum attainable height, in ft.
Kenneth S.
when the ball reaches ground level, s(t) = 0; the solution to the quadratic gives the time to fall to ground assuming that its path takes it wide of the building). The quadratic turns out not to be factorable, so you have to use the quadratic formula to find the zeros (roots).
To answer b, set s(t) = 100 & solve for t.
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01/30/17
Ashley N.
so I got a) the quadratic formula is the best because I cannot factor the problem
b) I got 4.78 seconds
so for c) do I just set s(t) to equal 0 and then use the quadratic formula?
And for d) do I just set s(t) to equal 300?
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01/30/17
Kenneth S.
Ashley, read part "d)" answer that I gave originally. If highest point (vertex of upside-down parabola) is maximum s value as I indicated, then clearly 300 ft. height cannot be attained.
Because of the (poor) way that this question was written (given to you), we already have the answer for "c;" I gave this as "a)" in my answer. Actually, in THEIR "a)", there was nothing to solve--I went ahead and set s(t) = 0 and thus anticipated their "c" question.
When you're all done with this work, review it from the beginning and strengthen you understanding of the whole deal.
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01/30/17
Ashley N.
Thsnk you very much for your time! I was super confused due to the way this was written. I think I understand most of it now. I know that obviously the ball can never reach 300 feet but I'm supposed to show why it doesn't reach 300ft algebraically. I set s(t) to equal 300 which gave me two negative numbers numbers which shows that it's impossible. And just so I'm sure I understand what you're saying (because my math vocabulary and skills overall aren't amazing) for c) I just need to set s(t) to equal 0 and then solve using the quadratic formula? I was just confused about what surface my teacher may have been talking about but I guess she means ground surface then? Thanks again! Sorry for so many questions, math just isn't my stong point.
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01/31/17
Kenneth S.
You are very welcome. I am finding that many math questions nowadays are poorly written, and this "surface" ambiguity is one of them. I attribute this phenomenon to the forced transfer from older well vetted textbooks to recent for-profit Common Core materials.
Showing that the vertex, highest point on path, has s(t) < 300 ft is a valid mathematical step--based on reasoning and the formula for t-coordinate of the vertex. The other way, using s(t) = 300 & obtaining both t roots < 0 is an alternative.
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01/31/17
Ashley N.
01/30/17