1. A horizontal spring, of constant 12N/m, is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 x 10^-3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor? 2. Relevant equations E = 0.5kx^2 E = 0.5mv^2 delta d = 0.5(acceleration)(delta time)^2 dx = vx x t 3. The attempt at a solution E = 0.5kx^2 = 0.5(12N/m)(0.04m)^2 = 0.0096J E = 0.5mv^2 0.0096J = 0.5(8.3x10^-3kg)(v)^2 Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v 2.31 m/s = v delta d = 0.5at^2 t = square root of [ (2 times delta d) / a] *accleration = gravity, 9.8m/s t = square root of [ (2 x 0.93m) / (9.8m/s^2)] t = 0.436s dx = vx x t = 2.31 m/s x 0.436s = 1.00716m I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance!