if a person shoots a basketball overhand from a position 8 feet above the floor then the path can sometimes be modeled by the quadratic function defined by
f(x)= -16x^2/0.434v^2+1.15x+8
where v is the initial velocity of the ball in feet per second...so if the hoop is 10 feet high and 15 feet away what initial velocity does the basketball have?

- So I did what the online homework said to do and subed 15 for x and put it in the graphing calc and then put y= 10 in the graph calc and found the intersect...which is 0.044 but I can not for the life of me get the velocity

Totally feel like I am missing something critical.
Any help is appreciated.

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Any idea where this formula comes from? I didn't try to derive the formula, but I'll accept it as a given, since I gather it is part of the problem. I guess buried in it are things like the angle at which the shooter throws the ball.

Did you miss a set of parenthesis. Do you think the formula is actually:
>> y = f(x)= -16 x²/(0.434v²) + 1.15x + 8

>> I did what the online homework said to do and subed 15 for x ... put y= 10

So, is this what you did?

10 = -16 (15)²/(0.434v²) + 1.15(15) + 8

All I see here are numbers and v² which is solvable.

Your equation is a mess
but given that there maybe dimensions hidden in the constants and the it is a very special equation ,note that at x = 15 f(x) should be 10 because the ball should be at the height of the basket. putting x = 15 and f(x) = 10 the only unknown in your equation is v. solve for v.

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Notice that y = y(x) has only two terms. Your formula has three terms. The last term in your formula is +8, which is the height of the ball (i.e., 8 feet above the floor - must be a tall shooter).

That is, f(0) = 8 for your formula. But the formula in the link is showing an intial height of 0 (i.e., f(0) = 0), so that term is left out in the link.

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