If a person shoots a basketball overhand from a position of 8 feet from 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. If the basketball hoop is 10 feet high and located 17 feet away, what initial velocity v should the basketball have to go through the hoop?

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

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Tringa0 Tringa0 · Answer 2 · 2019-10-17T07:59:26+02:00

Answer:

The initial velocity of ball should be 0.0406 ft/s.

Step-by-step explanation:

[tex]y=\frac{-16x^2}{0.434v^2}+1.15x+8[/tex]

x = the distance away from the hoop

y = Height of the hoop

v = velocity of the ball

Given : x = 17 ft, y = 10 ft , v = ?

[tex]10=\frac{-16(17)^2}{0.434v^2}+1.15\times (17)+8[/tex]

[tex]10-8-19.55=\frac{-16(17)^2}{0.434v^2}[/tex]

[tex]\frac{-17.55\times 0.434}{-16\times (17)^2}=v^2[/tex]

[tex]v^2=0.001647[/tex]

v = 0.0406 ft/s

The initial velocity of ball should be 0.0406 ft/s.