Question

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?

SIDE NOTE: I'm having trouble solving this graphically AND mathematically.

Answer 1

I don't know I was never that good at basketball.

Answer 2

this might help if you post this into the physics department but i solved a similar question today using the projectile forumla and looks like you can use the same equation \[y = x \tan(\alpha) - \frac{ g -x^2 }{ 2-v_{0}^2 } *(1 + tg^2 \alpha)\]

Answer 3

so you have x = 17, and y = 10, put them values in, and solve for v

Answer 4

Yes, when I tried isolating v, I messed up somewhere. My homework (online) is insisting I solve it graphically, but when I graph the equations y=10 and y=(-4624/0.434x^2)+27.55, my intersect point isn't the same one the online module comes up with. I can't figure out how I might be entering the equation wrong.

Answer 5

hmm, Sorry :(

Answer 6

You should get \[

v\approx24.64

\]

Answer 7

A reasonable speed

Answer 8

My calculator is giving me a weird intersect point of -.0405858, but the online homework module is saying it's approx. 24.64.

I can't figure out where I'm messing up!

Answer 9

Could either one of you walk me through isolating v?

Answer 10

first you get
\[
10=27.55\, -\frac{10654.4}{v^2}\\
\frac{10654.4}{v^2}=27.55\, -10= 17.55
\]

Answer 11

Yeah, you just need the parabola to go through (17,10) so
\[10 = \frac{-16(17)^2}{0.434v^2} + 1.15(17) + 8\]
Subtract 8 from both sides
\[2 = \frac{-16(17)^2}{0.434v^2} + 1.15(17)\]Subtract 1.15(17) from both sides\]
\[2-1.15(17) = \frac{-16(17)^2}{0.434v^2}\]Multiply both sides by \(0.434v^2\)
\[0.434v^2(2-1.15(17)) = -16(17)^2\]Divide both sides by 0.434(2-1.15(17))
\[v^2=\frac{-16(17)^2}{0.434(2-1.15(17))}\]Take square root of both sides
\[v = \sqrt{\frac{-16(17)^2}{0.434(2-1.15(17))}}\]

Answer 12

\[
10=27.55\, -\frac{10654.4}{v^2}\\
\frac{10654.4}{v^2}=27.55\, -10= 17.55\\
v^2= \frac {10654.4}{17.55}=607.088\\
v= 24.6392
\]

Answer 13

Here's a graph: the horizontal line represents the height of the hoop, and you can see it crosses the curve at x = 17

Answer 14

How did you enter the equation into the calculator? I'm getting a line whose parabola hits the x axis, not at y=10.

Answer 15

Well, I'll tell you, but it won't do you much good :-)

Plot[{10, (-16 x^2)/(0.434 (24.6391)^2) + 1.15 (x) + 8}, {x, 0, 25},
AspectRatio -> Automatic, PlotRange -> {{0, 25}, {0, 15}}]

Answer 16

the equation itself is (-16 x^2)/(0.434 (24.6391)^2) + 1.15 (x) + 8

Answer 17

Yeah, the homework module was telling me to graph y=10 and then y=(-16x^2)/(0.434x^2)+19.55+8

Answer 18

subbing x for v

Answer 19

but the parabola does hit the x-axis, of course — it's just at different points. I also graphed y = 10, that was the horizontal blue line.

Answer 20

oh, you don't sub x for v...x and v are independent! you get a different parabola for each different value of v — that's the speed of the ball.

Answer 21

No, no, I just mean for the purposes of entering the equation into the graph

Answer 22

there's no v variable..so I used x instead

(-4624)/(0.434x^2)-19.55+8

Answer 23

even then. you only get one variable, which is x. you have to enter the value you found for v in place of v.

here's a graph with 3 different curves representing our true solution, and v = 23 and v = 26

Answer 24

but x is in the numerator! the number in the denominator is constant for any given parabola of interest. the formula gives us the height of the ball at position x.

Answer 25

Hrm..okay. But the problem was trying to get me to solve it solely by graphing those two equations. That's why I was having such a hard time. The vertex of the graph I was getting ended at the x axis.

Answer 26

well, you can get it graphically by graphing the equation using x = 17, and picking different versions of v until you find the one that hits the target. but that's a tedious iterative procedure (especially if you don't have a computer to do the graphing for you!)

Answer 27

or you can just solve the parabola, as we did earlier.

Answer 28

Yeah, just my TI-83.

Answer 29

Well, thank you! You've both been VERY helpful.