# For which angle is the speed of a basketball more important to score?

1. Nov 18, 2009

### imatreyu

1. The problem statement, all variables and given/known data

"A 3-point jump shot is release 2.2 m above the ground, 6.02 m from the basket, which is 3.05 m high. For launch angles of 30 degrees and 60 degrees find the speed needed to make the basket."

For which angle is it more important that they player get the speed right? To explore this question, vary the speed at each angle by 5% and find the change in the range of the throw.

2. Relevant equations

the range formula? (but since the starting and ending heights aren't the same, this must be incorrect.)

3. The attempt at a solution

I already have found the speeds needed to make the basket.

9.5 m/s for the 30 degree angle and 8.6 m/s for the 60 degree angle.

I varied each one by 5 percent (above and below), but have no clue about where to go from there. Where do these values go?

I know that the answer is that the high launch angle is less sensitive to speed variations, and thus speed must be more important at the lower launch angle.

Thank you!

2. Nov 18, 2009

### clamtrox

If you vary the correct speed of the throw by 5%, how much does the distance 6.02 m change? So you just have to work your way back with the same equations.

3. Nov 18, 2009

### imatreyu

Thank you! Haha. . .So it's just math. The problem is, this is very simple math, but I'm completely lost as for how to solve here for x:

y (change) = x tan(theta) - ((ax^2)/ 2vi^2 (cos theta)^2)

Can it possibly become quadratic?

4. Nov 18, 2009

### HallsofIvy

What do you mean "become" quadratic? That is quadratic- it has the x^2 right there. That is basically y= Bx- Ax^2 with B= tan(theta) and A= a/(2vi^2(cos theta)^2) which is, of course, the same as Ax^2- Bx+ y= 0.

5. Nov 18, 2009

### imatreyu

Yes!

Thank you!

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