1. The problem statement, all variables and given/known data 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) 2. Relevant equations r2y = r1y + v1y(t2) - 4.9(t22) same for x-axis analysis 3. 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.