Vx = 10cos(30) You need to find t, the time it takes to hit the ground. You do this by finding Vy (10sin(30)), and using a = -9.81, and an additional height of 10m, solve for t. Using x = .5a(t^2) + v_0(t) + x_0 x_vertical = .5(-9.81)(t^2) + (10sin(30))t + 10 = 0 x_horizontal = .5(0)(t^2) + (10cos(30))t + 0 Do you understand why acceleration for the horizontal distance is zero? [Edited on 11.15.2012 11:38 PM PST]