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Mathematics 18 Online
OpenStudy (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

OpenStudy (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

OpenStudy (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.

OpenStudy (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?

OpenStudy (ohohaye):

@ybarrap @mathmale

OpenStudy (ohohaye):

@ParthKohli can you please help me

jimthompson5910 (jim_thompson5910):

think of `h = -5t^2 + 30t + 10` as `y = -5x^2 + 30x + 10`

jimthompson5910 (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?

jimthompson5910 (jim_thompson5910):

any ideas @ohohaye ?

OpenStudy (ohohaye):

a=5 b=30 c=10

jimthompson5910 (jim_thompson5910):

a = -5 actually

OpenStudy (ohohaye):

Oops, didn't see the negitive sign

jimthompson5910 (jim_thompson5910):

now you need to use this formula \[\Large h = -\frac{b}{2a}\]

jimthompson5910 (jim_thompson5910):

\[\Large h = -\frac{b}{2a}\] \[\Large h = -\frac{30}{2(-5)}\] \[\Large h = ???\]

OpenStudy (ohohaye):

h=-3

jimthompson5910 (jim_thompson5910):

the two negatives will cancel to form a positive

OpenStudy (ohohaye):

Oh ok, so it's just 3?

jimthompson5910 (jim_thompson5910):

yes, h = 3 is where the rocket reaches the peak height |dw:1450242441052:dw|

OpenStudy (ohohaye):

Ok, so I would use the \[h=-\frac{ b }{ 2(a) }\] if I had a question like this on a test or something?

jimthompson5910 (jim_thompson5910):

yes that will get you the x coordinate of the vertex

OpenStudy (ohohaye):

I used this to find the max height of the rocket though

jimthompson5910 (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

jimthompson5910 (jim_thompson5910):

to find the max height, you plug t = 3 back into the original equation

OpenStudy (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?

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

what happens when you plug in t = 3 ?

OpenStudy (ohohaye):

I get 55

jimthompson5910 (jim_thompson5910):

so 55 is your max height

OpenStudy (ohohaye):

Yep

jimthompson5910 (jim_thompson5910):

|dw:1450243012057:dw|

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