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?

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

Step-by-step explanation:

dude i am so late sorry bro uh well i hope i help SOMEONE

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question