ohohaye
  • ohohaye
How long will it take for the rocket to reach maximum height? Equation: h= -5t^2+30t+10 Please show me how to do this
Mathematics
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SOLVED
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katieb
  • katieb
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ybarrap
  • ybarrap
Take derivative, set equal to zero, solve for t. Why does this work? When you set the derivative equal to zero, you are asking the question, when is the function h flat (i.e. has zero slope)? The function h changes direction and goes from positive slope to negative slope at the peak of the flight path, which is at the apex of the parabola. When something goes from positive to negative, it must go through zero. Hence, setting derivative of h to zero is asking when is the slope zero - this locates the apex of h. Make sense? http://www.wolframalpha.com/input/?i=d%2Fdt+%28+-5t%5E2%2B30t%2B10%29+%3D+0 http://www.wolframalpha.com/input/?i=h%3D+-5t%5E2%2B30t%2B10
mathmale
  • mathmale
An alternative approach, perhaps shorter but not as exciting mathematically, would be to find the vertex of the given quadratic function. The first coordinate of this vertex would be the time at which the rocket reaches its max. height.
ohohaye
  • ohohaye
So finding the time at which the rocket hits the max height is the same way to find the time in which the rocket would land?

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ohohaye
  • ohohaye
@ybarrap @mathmale
ohohaye
  • ohohaye
@ParthKohli can you please help me
jim_thompson5910
  • jim_thompson5910
think of `h = -5t^2 + 30t + 10` as `y = -5x^2 + 30x + 10`
jim_thompson5910
  • jim_thompson5910
`y = -5x^2 + 30x + 10` is in the form `y = ax^2 + bx + c` can you tell me what 'a', 'b', & 'c' are?
jim_thompson5910
  • jim_thompson5910
any ideas @ohohaye ?
ohohaye
  • ohohaye
a=5 b=30 c=10
jim_thompson5910
  • jim_thompson5910
a = -5 actually
ohohaye
  • ohohaye
Oops, didn't see the negitive sign
jim_thompson5910
  • jim_thompson5910
now you need to use this formula \[\Large h = -\frac{b}{2a}\]
jim_thompson5910
  • jim_thompson5910
\[\Large h = -\frac{b}{2a}\] \[\Large h = -\frac{30}{2(-5)}\] \[\Large h = ???\]
ohohaye
  • ohohaye
h=-3
jim_thompson5910
  • jim_thompson5910
the two negatives will cancel to form a positive
ohohaye
  • ohohaye
Oh ok, so it's just 3?
jim_thompson5910
  • jim_thompson5910
yes, h = 3 is where the rocket reaches the peak height |dw:1450242441052:dw|
ohohaye
  • ohohaye
Ok, so I would use the \[h=-\frac{ b }{ 2(a) }\] if I had a question like this on a test or something?
jim_thompson5910
  • jim_thompson5910
yes that will get you the x coordinate of the vertex
ohohaye
  • ohohaye
I used this to find the max height of the rocket though
jim_thompson5910
  • jim_thompson5910
in this case, t takes place of x so if we know the x coordinate of the vertex, then we know how long it takes til the object reaches the peak
jim_thompson5910
  • jim_thompson5910
to find the max height, you plug t = 3 back into the original equation
ohohaye
  • ohohaye
Oh, so to find the time, I only have to use the first formula, but the find the max height I have to plug the answer back into the equation?
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
what happens when you plug in t = 3 ?
ohohaye
  • ohohaye
I get 55
jim_thompson5910
  • jim_thompson5910
so 55 is your max height
ohohaye
  • ohohaye
Yep
jim_thompson5910
  • jim_thompson5910
|dw:1450243012057:dw|
ohohaye
  • ohohaye
Ok, thank you
jim_thompson5910
  • jim_thompson5910
you're welcome
ohohaye
  • ohohaye
Could you help me with another problem
jim_thompson5910
  • jim_thompson5910
sure
ohohaye
  • ohohaye
Should I start a new question and add you or just use this one?
jim_thompson5910
  • jim_thompson5910
a new question is preferable to avoid clutter and lag
jim_thompson5910
  • jim_thompson5910
tag me in the new question
ohohaye
  • ohohaye
Ok

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