1. The problem statement, all variables and given/known data A basketball player shoots the ball at a height of 1.7m at an angle of 60 degrees into a hoop. The range is 10ft or 3.05m. The height from the ground to the top of the hoop is 10ft or 3.05m. What is the initial velocity m/s to make it to the hoop? 2. Relevant equations 1) Vy^2 = Viy^2 - 2g(Change in Y) 2) Y = Yi + Viy(t) - 1/2gt^2 3) Vy = Viy - gt 4) X = Xi + Vixt = (ViCos(theta)) t 3. The attempt at a solution B/c the ball is shot at 1.7m off the ground I reduced this to zero thereby reducing the height of the hoop by 1.7m. So I have the following givens: x= 3.05m y = 3.05m - 1.7m = 1.35m Theta = 60 degrees Xi = 0 m Yi = 0 m Time is unknown Solving for Initial Velocity (Vi) I am assuming max height is enough to clear the hoop at 1.35m. So at max height Velocity ='s 0 m/s. So now I used X = Xi + Vixt = ViCos(Theta)t 3.05 = 0 + Vi (Cos 60) t t = 6.1 / Vi Now I take t and plug into y = Yi + Viyt - 1/2gt^2 1.35m = 0 + Visin(60)t -1/2(9.8)(t^2) 1.35 = (0.866025Vi)(6.1/Vi) - 4.9 (6.1/Vi)^2 1.35 = 5.282754 - 182.329/Vi^2 -3.932754 = -182.329 / Vi^2 Vi^2 = 46.36165903 Vi = 6.8089 m / s Does that seem right or am I not taking into account other factors. Thanks.