Discover the Solution to a Challenging Basketball Physics Problem

AI Thread Summary
A basketball player needs to determine the initial speed required to shoot a ball through a hoop 3.05 m high from a distance of 13.0 m, at a 43.0-degree angle. The player expresses confusion about how to apply kinematic equations due to insufficient information on variables. Key variables include the horizontal distance (∆x = 13.0 m) and the vertical distance (∆y = 1.05 m), which is the difference in height between the player's position and the hoop. The discussion emphasizes the need to eliminate time from the equations to derive a quadratic equation for the initial speed. The conversation highlights the importance of understanding projectile motion and the correct application of kinematic formulas.
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
 
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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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