Solving for an angle in 2-D kinematics

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Homework Help Overview

The problem involves a basketball player attempting a shot from a distance of 14 meters, with the ball leaving her hands 1.4 meters below the basket. The player throws the ball at a speed of 18 m/s, and the goal is to determine the necessary angle for the shot to succeed.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations for both vertical and horizontal motion, with attempts to express time in terms of angle. There are inquiries about trigonometric identities relating tanθ and secθ, and suggestions to reformulate the problem using a substitution for tanθ.

Discussion Status

The discussion has progressed with participants exploring various mathematical approaches and identities. Some guidance has been offered regarding the formulation of a quadratic equation, which has led to a clearer path for one participant. However, there is still an ongoing exploration of the relationships between the trigonometric functions involved.

Contextual Notes

Participants note challenges with combining terms involving tanθ and secθ, as well as the lack of prior experience with certain mathematical techniques. There is an acknowledgment of the complexity of the problem and the need for further clarification on the relationships between the variables.

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Homework Statement


At the buzzer, a basketball player shoots a desperation shot. She is 14m from the basket and the ball leaves her hands exactly 1.4m below the rim. She throws the ball at 18m/s. Can she make the shot? (Solve for necessary angle)

Homework Equations



r2y = r1y + v1y(t2) - 4.9(t22)

same for x-axis analysis

The Attempt at a Solution



For y-position (using above equation):
1.4m = 18(m/s)sinθ(t2s) - 4.9(m/s2)(t2s)2
1.4m = 18(t2)(sinθ)(m) - 4.9(t22)(m)

For x-position:
14m = 18(cosθ)(t2)
t2 ≈ 0.78/(cosθ)

After this, I'm not sure what to do. When I plug the t2 value I got solving the x-position into the y-position equation, I end up with a tanθ as well as a sec2θ with no visible way to combine or cancel the two. I also don't see any way to substitute for the trig values to get them to cancel each other in any way. Any help would be greatly appreciated.
 
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Hello.

Do you know a trig identity that relates tanθ and secθ?
 
TSny said:
Hello.

Do you know a trig identity that relates tanθ and secθ?

Well, I did try that, but then I ended up with a tanθ with a value minus a tan2θ with a value in the equation, which once again left me with not knowing what to do. If I try to substitute the tan part, then I end up with a square root and once again no way for them to cancel one another or combine.
 
Suppose you let z = tanθ. Can you write your equation as a quadratic equation in z?
 
TSny said:
Suppose you let z = tanθ. Can you write your equation as a quadratic equation in z?

Ok, now I see where you are going with that. Thanks a lot. We never did anything like that before, but from what you are saying, I think I can solve it now.
 
Just so you know, I did what you told me and got the angles right. It could be either 18 or 77 degrees. Thanks so much for your help on this one. Although there was no way I was going to figure out I could do that on my own, as soon as you mentioned it, it made perfect sense.
 
OK, good work!
 

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