1. The problem statement, all variables and given/known data A golfer strikes a golf ball at an angle of 17 degrees above the horizontal. With what velocity must the ball be hit to reach the green which is a horizontal distance of 250 m from the golfer? It is at the same height as the tee. 2. Relevant equations trig equations v2^2=v1^2+2ad d=v1(t)+1/2a(t)2 d=(v1+v2/2)t 3. The attempt at a solution Well I used trigonometry to find the dy (tan17*250) and I believe that's the maximum height.......meaning v2y at that time is 0 So I have dy=76.43 m a=-9.8m/s^2 v2y=0 I used v2^2=v1^2+2ad to find the initial velocity in the y direction and got 38.7 I plugged that into the equation d=(v1+v2/2)t to find time and got 1.975 seconds For the x direction I know dx=250, a=0, and t=1.975 seconds So I used d=v1(t)+1/2a(t)2 to find v1 but the a is 0 so the equation just ends up being d=v1(t) and I got the initial velocity in the x direction to be 126.59 I used pythagorean theorem to find the total velocity to be 132.34 m/s 17 degrees above the horizontal....but the answer is 66m/s 17 degrees above the horizontal.