Question 654288: If an object is thrown upward with an initial velocity of 96ft/sec,its height after t seconds is given by h(t)=96t-16tsquared. Find the maximum height of the object.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! If an object is thrown upward with an initial velocity of 96ft/sec,its height after t seconds is given by h(t)=96t-16tsquared. Find the maximum height of the object.
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h(t)=96t-16tsquared
h(t)=96t-16t^2
h(t)=-16t^2+96t
This is an equation of a parabola that opens downwards.
Its standard form: y=(x-h)^2+k, (h,k) = (x,y) coordinates of the vertex.
maximum=y-coordinate of vertex.
..
complete the square:
h(t)=-16(t^2-6t+9)+144=-16(t-3)^2+144
vertex: (3,144)
maximum height of the object=144 ft
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