Discover the Solution to a Challenging Basketball Physics Problem

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Homework Help Overview

The problem involves a basketball player attempting to determine the initial speed required to shoot a basketball through a hoop, given specific heights and distances. The context is rooted in kinematics and projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations in both horizontal and vertical directions, with some expressing uncertainty about the available information and how to derive necessary variables. Questions arise regarding the interpretation of delta x and delta y based on the problem's diagram.

Discussion Status

Some participants have offered guidance on using kinematic equations and eliminating time to form a quadratic equation. However, there is ongoing uncertainty about the correct application of these equations and the interpretation of the problem's parameters.

Contextual Notes

Participants note confusion regarding the use of certain variables, particularly delta y and delta x, and the implications of the basketball's trajectory. There is a mention of the diagram not providing clear information for computing these values.

intriqet
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Homework Statement



A basketball player who is 2.0 m tall is standing on the floor L = 13.0 m from the basket. If he shoots the ball at a 43.0deg angle with the horizontal, at what initial speed must he throw so that the ball goes through the hoop without striking the backboard? The basketball height is 3.05 m.


Homework Equations



Kinematic equations.


The Attempt at a Solution



I don't even know where to start with this problem. I feel like I hadn't been given enough information. Please post hints to get me started. Thanks
 
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Hi intriqet! :wink:

Let the initial speed be vi, and the time taken be t, and use the standard constant acceleration equations, once for the x direction (with a = 0), and once for the y direction (with a = -g).

Eliminate t, and that should give you a quadratic equation for vi. :smile:
 
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I'm not sure how to apply the constant acceleration kinematic formulas because I only have information for two of the variables for each direction.

For instance I have for the x (horizontal) direction:
Ax = 0
Vxi = xcos43

and for the y:
Ay = -9.81
Vyi = xsin43

If I had delta x or y this would be a ton easier but I can't compute delta x or delta y from the diagram.
 
intriqet said:
If I had delta x or y this would be a ton easier but I can't compute delta x or delta y from the diagram.

uhh? :confused:

from the diagram, ∆x = L = 13.0, and ∆y = 1.05.
 
Sorry I wasnt sure If I could use delta y = 1.05 because that isn't the max height of the parabola and delta x = 13 because the full length of the parabola isn't 13.

Ok so based on that information I derived that

Ax = 0
Vxi = xcos43
Vxf = xcos43
deltax = 13
tx = ty

Ay = -9.81
Vyi = xsin43
Vyf = sqrt(-20.601 + (xsin43)^2)
deltaY = 1.05
ty = tx

Is this correct so far?

Oh and thanks by the way for your help and input
 
Last edited:
intriqet said:
Sorry I wasnt sure If I could use delta y = 1.05 because that isn't the max height of the parabola and delta x = 13 because the full length of the parabola isn't 13.

Forget parabola!

(and anyway, the height and width of a parabola are infinite)
 

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