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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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few ways to solve this
are you familiar with derivatives?
with a graph correct?

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Other answers:

Without Calculus y|dw:1349657074734:dw|ou know that this graph will look like this
oh ok
so you know it hits its maximum height at its' vertex. you know how to find the x for the vertex by \[x=\frac{-b}{2a}\]
x so happens to be time in our case =P
21 seconds
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
without calculus \[ax^2+bx+c\] in our case \[ax^2+bx\] \[x=\frac{-b}{2a}\] \[\frac{-440}{-32}\] 13.75 seconds
oh wow i was completly of thanks alot !!!

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