1. The problem statement, all variables and given/known data A ball is kicked from the ground at an angle of 65° and at a speed of 8 m/s. What is its speed when it reaches 1/2 of the maximum height along the path, assuming that air resistance is negligible? My head hurts from this. I made a diagram of this event by using a vector pointing diagonally with 65 degrees to the horizontal I used trig to find the velocity at the x and y components sin65 = y/8 Y = 7.25m/s cos65=x/8 x = 3.38m/s from here, i didn't know what to do. I used the equation V^2 = Vi^2+2a[tex]\Delta[/tex]x [tex]\Delta[/tex]x = .58287751m Then, to obtain the height tang65=H/.58287751m H = 1.25m then half of that is .6249m and uh....conservation of energy Intial 1/2mVi^2 = Final mgh + 1/2mVf^2 You can cancel out the m's 1/2(8)^2 = (9.80)(.6249m) + .2Vf^2 Vf= 7.19 = 7.2m/s ?