A basketball of mass 0.624 kg is shot from a vertical height of 1.55 m and at a speed of 12.1 m/s. After reaching its maximum height, the ball moves into the hoop on its downward path, at 3.05 m above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop. v=√(v0^2+2g(y0-y)) I've tried plugging the numbers in an I come up with an answer of 13.26 m/s. However this is incorrect. I believe I might be thinking of where the ball is being shot from differently. v=√((12.1^2)+2(9.81)(1.55-3.05) Any help would be appreciated. Thanks, Dominic.