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anonymous
 3 years ago
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?
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0few ways to solve this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you familiar with derivatives?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0with a graph correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Without Calculus ydw:1349657074734:dwou know that this graph will look like this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you know it hits its maximum height at its' vertex. you know how to find the x for the vertex by \[x=\frac{b}{2a}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x so happens to be time in our case =P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1349657191197:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0with 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0without calculus \[ax^2+bx+c\] in our case \[ax^2+bx\] \[x=\frac{b}{2a}\] \[\frac{440}{32}\] 13.75 seconds

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wow i was completly of thanks alot !!!
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