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

AI Thread Summary
In the discussion, the focus is on determining the importance of speed for scoring a basketball from different launch angles, specifically 30 degrees and 60 degrees. The required speeds to make the basket are identified as 9.5 m/s for the lower angle and 8.6 m/s for the higher angle. It is concluded that the lower launch angle is more sensitive to speed variations, making precise speed more critical for successful shots at that angle. The participants also explore the mathematical equations involved in calculating the change in distance when varying speed by 5%. Ultimately, the discussion emphasizes the relationship between launch angle, speed, and scoring accuracy in basketball.
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Homework Statement



"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.



Homework Equations



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

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!
 
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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.
 
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?
 
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.
 
Yes!

Thank you!
 
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