If a person shoots a basketball overhand from a position 8 feet above the​ floor, then the path of the basketball through the hoop can be modeled by the parabola y equals StartFraction negative 16 x squared Over 0.434 v squared EndFraction plus 1.15 x plus 8y=−16x20.434v2+1.15x+8​, where v is the velocity of the ball in​ ft/sec, y is the height of the hoop and x is the distance away from the hoop. ​(a) If the basketball hoop is 10 ft high and located 1212 ft​ away, what initial velocity v should the basketball​ have? ​(b) What is the maximum height of the​ basketball?

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A: Topic = Application of Derivative

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A: See the attachment

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A: Introduction ::

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Q: What is the maximum vertical distance between the line y = x + 6 and the parabola y = x< for -2 <x<…

A: To Find

Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.