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A projectile is thrown upward so that its distance above the ground after t sec is given by h(t)=-11t²+264t. After how many sec...Asked by Kristina
A projectile is thrown upward so that its distance above the ground after t seconds is h(t)=-16t^2 + 570t. After how many seconds does it reach its maximum height?
Answers
Answered by
Damon
570 feet/second ????
You threw it up at about 400 miles per hour, typo but anyway
where is the vertex of that parabola?
16 t^2 -570 t = -h
t^2 - 265/8 t +265^2/16^2 = -h+265^2/16^2
(t-265/16)^2 = etc
so vertex at t = 265/16
BUT
I do not believe you
You threw it up at about 400 miles per hour, typo but anyway
where is the vertex of that parabola?
16 t^2 -570 t = -h
t^2 - 265/8 t +265^2/16^2 = -h+265^2/16^2
(t-265/16)^2 = etc
so vertex at t = 265/16
BUT
I do not believe you
Answered by
Steve
vertex at t = -b/2a = 570/32 = 285/16
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