Question

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=(-16x^2)/(0.434v^2)+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 1717 ft​ away, what initial velocity v should the basketball​ have?
​(b) What is the maximum height of the​ basketball?

Answer 1

what is the maximum height of the basketball

Answer 2

The ath of the basketball through the hoop can be modeled by the parabola y=(-16x^2/0.434v^2)+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. If the basketball hoop is 10 feet high and located 17 feet away then:

y = (-16x^2/0.434v^2)+1.15x+8

10 = [-16(17)^2/0.434v^2] + 1.15(17) + 8

v = 24.64 ft/s

Explanation:

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