1. The problem statement, all variables and given/known data A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a θ = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m. How much time has elapsed between when the block is released and when it hits the floor? 2. Relevant equations 3. The attempt at a solution I was required to find the acceleration of the block as it moves down the incline. I found that to be 5.621 m/s^2 I then found the velocity at the point where the block leaves the incline to be 3.13 m/s. Then I found the distance that the block travels from when it leaves the table to when it hits the ground. This I found to be 1.234 m. Then I attempted to find the total time of the blocks movement: Velocity in the y direction: Vyo = 3.13 Cos(55) Vyo = -1.7953 m/s Time for the block from the edge of the table to the floor: Y = Yo + Vyo - 1/2at^2 0 = 2m - 1.7953t - 1/2(9.8)t^2 t = .48143 sec Velocity in the x direction: Vxo = 3.13 Sin(55) Vxo = 2.56395 m/s Vf = Vo + at 2.56395 = 0 + 5.621(t) t = .456 sec So, the total time would be the two times added together: .48143 sec + .456 sec = .93743 sec I submitted this answer and it said it was within 10% of the correct answer....Am I making some sort of rounding error or is there some mistake in my calculations? Thanks for any help!