anonymous
  • anonymous
Derek kicks a soccer ball off the ground and in the air with an initial velocity of 31 feet per second. Using the formula H(t) = -16t2 + vt + s, what is the maximum height the soccer ball reaches?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jdoe0001
  • jdoe0001
well keep in mind that \(\bf \text{initial velocity}\\ h = -16t^2+v_ot+h_o \text{ , in feet}\\ v_o = \textit{initial velocity}\\ h_o = \textit{initial height}\\ h = \textit{height at "t" seconds}\)
jdoe0001
  • jdoe0001
so plugging in the data available it says that the ball got kicked "off the ground", so the initial height =0 the initial velocity is given, 31 ft/s so our equation will look like \(\bf H(t) = -16t2 + 31t + 0\)
anonymous
  • anonymous
so solve for t

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jdoe0001
  • jdoe0001
notice that the equation is a parabola.... well \(\bf H(t) = -16t^2 + 31t + 0\) it has a negative leading coefficient, meaning that it's opening downwards like |dw:1377715444670:dw|
anonymous
  • anonymous
ok. whats a coefficient?
jdoe0001
  • jdoe0001
well, no, solving for "t" will give us the "x-intercepts" or when the ball is at the ground we want to know that maximum of the parabola, that is, the "hump" point or the vertex point, keep in mind that usually the y-axis represents the Height, the x-axis represents the Seconds so we want the y-axis value at the vertex point for a parabola like so, you can always find it's vertex at \(\bf \left(-\cfrac{b}{2a}, c-\cfrac{b^2}{4a}\right)\)
jdoe0001
  • jdoe0001
http://www.mathsisfun.com/definitions/coefficient.html
anonymous
  • anonymous
do you have an answer because i am going to do the math let me know if i got it right?
jdoe0001
  • jdoe0001
no I don't, but I can get it from \(\bf \left(-\cfrac{b}{2a}, c-\cfrac{b^2}{4a}\right)\) quick enough
jdoe0001
  • jdoe0001
the y-axis component is the only one you need from the vertex, the y-axis is the Height at the vertex point
anonymous
  • anonymous
ok one second let me solve it
anonymous
  • anonymous
14.2 seconds is what i got
anonymous
  • anonymous
is it right?
jdoe0001
  • jdoe0001
well, \(\begin{matrix} &a&b&c\\ H(t) =& -16t^2& + 31t& + 0\\ &&&& \large c-\cfrac{b^2}{4a}\implies \large 0-\cfrac{31^2}{4(-16)} \end{matrix}\)
jdoe0001
  • jdoe0001
I got something a bit bigger
jdoe0001
  • jdoe0001
well, then again, I didn't calculate the "x", since that's seconds, but you're asked on Height, that's the "y-axis"
anonymous
  • anonymous
its something like 15. something or 14. something
jdoe0001
  • jdoe0001
so, that's your value, that's how high the ball gets
anonymous
  • anonymous
15 or 14?
jdoe0001
  • jdoe0001
well, what did you get from \(\bf 0-\cfrac{31^2}{4(-16)} \quad ?\)
anonymous
  • anonymous
28.264705
jdoe0001
  • jdoe0001
recheck :/
anonymous
  • anonymous
31^2 over 64 = 31 times 31 = 961 over 64
jdoe0001
  • jdoe0001
yes, that's what I get, 961/64
anonymous
  • anonymous
divide 961 bye 64 equals 15.01
anonymous
  • anonymous
yay!
jdoe0001
  • jdoe0001
yeap => http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItMTZ4XjIrMzF4KzAiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItNi42NTk5OTk5OTk5OTk5OTkiLCI2LjM0MDAwMDAwMDAwMDAwMSIsIjguMyIsIjE2LjI5OTk5OTk5OTk5OTk5NyJdfV0-

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