A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.1 m/s at an angle of 52.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. Initial velocity = 17.1sin52=17.1 x 0.788 =13.47 m/s. max height = 9.81 v^2=u^2 + 2as v^2=0+2*9.81*6.25 v= 11.07 m/s My answer is wrong.