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

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Homework Help Overview

The problem involves a basketball jump shot, specifically analyzing the required speed for different launch angles to successfully score. The scenario includes a release height of 2.2 m, a distance of 6.02 m from the basket, and a basket height of 3.05 m. The angles of interest are 30 degrees and 60 degrees, with a focus on how variations in speed affect scoring ability.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessary speeds for each angle and how to assess the impact of varying these speeds by 5%. There is a question about how to apply these variations to the range of the throw, and some participants express confusion about the mathematical relationships involved.

Discussion Status

Some participants have identified the speeds needed for each angle and are exploring the implications of varying these speeds. There is acknowledgment of the mathematical nature of the problem, with some guidance provided on the quadratic form of the equations involved. However, there is no explicit consensus on the next steps or resolution of the confusion regarding the calculations.

Contextual Notes

Participants are working under the constraints of differing launch angles and the need to account for varying heights in their calculations. The original poster mentions uncertainty about the application of the range formula due to the differing starting and ending heights.

imatreyu
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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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