babydoll332
A projectile is thrown upward so that its distance above the ground after t seconds is h = - 16t2 + 440t.
After how many seconds does it reach its maximum height?
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Outkast3r09
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few ways to solve this
completeidiot
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are you familiar with derivatives?
babydoll332
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with a graph correct?
Outkast3r09
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Without Calculus y|dw:1349657074734:dw|ou know that this graph will look like this
babydoll332
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oh ok
Outkast3r09
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so you know it hits its maximum height at its' vertex.
you know how to find the x for the vertex by
\[x=\frac{-b}{2a}\]
Outkast3r09
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x so happens to be time in our case =P
babydoll332
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ohhhh
Outkast3r09
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|dw:1349657191197:dw|
babydoll332
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21 seconds
Outkast3r09
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with calculus
\[y=-16t^2+440t\]
\[y'=-32t+440=velocity\]
we know that the velocity of y is 0 at the point of max height so
\[0=-32t+440\]
\[-440=-32t\]
\[\frac{-440}{-32}=t\]
13.75 seconds
Outkast3r09
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without calculus
\[ax^2+bx+c\]
in our case
\[ax^2+bx\]
\[x=\frac{-b}{2a}\]
\[\frac{-440}{-32}\]
13.75 seconds
babydoll332
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oh wow i was completly of thanks alot !!!