Basketball player is 2m tall. He shoots at 40 degrees over a distance of 10m into a basketball net that is 3.05m. Whats the initial velocity. d=10m Theta=40 degrees Vxi = Vi cos(40) Ax=0 xi=0 xf=10 Vyi= Vi sin(40) Ay=-9.80 yf=1.05 (difference between player and net) 1. First I wanted to solve for time Xf = Xi + Vxi*t + (1/2)Ax*(t^2) 10 = 0 + (Vi*cos(40))t + (1/2)(0) (t^2) 10 = (Vi*cos(40))t t = (Vi*cos(40))/ 10 2. Then, I'd use the Yf formula to plug my new time into to get Vi since Vyi= Vi sin(40) Yf = Yi + Vyi*t + (1/2)Ay*(t^2) 1.05=0 + (Vi*sin(40))t - (1/2)(-9.80)(t^2) ---then substituting t into the 1.05=0 + (Vi*sin(40))t + (1/2)(-9.80)(t^2) and after all the algebraic dust settled I got --- Vi = 7.159.... the problem is the book says 10.7 m/s. Where did I go wrong??