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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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Outkast3r09Best ResponseYou've already chosen the best response.1
few ways to solve this
 one year ago

completeidiotBest ResponseYou've already chosen the best response.0
are you familiar with derivatives?
 one year ago

babydoll332Best ResponseYou've already chosen the best response.0
with a graph correct?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
Without Calculus ydw:1349657074734:dwou know that this graph will look like this
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so you know it hits its maximum height at its' vertex. you know how to find the x for the vertex by \[x=\frac{b}{2a}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
x so happens to be time in our case =P
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
dw:1349657191197:dw
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
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
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
without calculus \[ax^2+bx+c\] in our case \[ax^2+bx\] \[x=\frac{b}{2a}\] \[\frac{440}{32}\] 13.75 seconds
 one year ago

babydoll332Best ResponseYou've already chosen the best response.0
oh wow i was completly of thanks alot !!!
 one year ago
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