Solving for an angle in 2-D kinematics

[Ascendant78]

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

r_2y = r_1y + v_1y(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.

 

[TSny]

Hello.

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

 

[Ascendant78]

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.

 

[TSny]

Suppose you let z = tanθ. Can you write your equation as a quadratic equation in z?

 

[Ascendant78]

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.

 

[Ascendant78]

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.

 

[TSny]

OK, good work!