In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor 1.40 m from the base of the counter. If the height of the counter is .860 m, (a) with what velocity did the mug leave the counter, and (b) what was the direction of the mug's velocity just before it hit the floor? (a) I have an equation, h = (vi^2(sinθi)^2)/2g, but I think it's wiser to use the equation of the vertical component of rf = ri + vit + .5a(t^2), which is yf = yi + vyit + .5a(t^2). The problem deals with the portion of the trajectory that has a negative velocity. For this portion, the final y coordinate is 0, the initial y coordinate is .860, and the initial velocity of y is 0, because it "starts" at the maximum height. Gravity is the driving force, and that is -9.8. So: yf = yi + vyit + .5a(t^2) 0 = .860 + (0 * t) + .5(9.8)(t^2) t = .41893938 seconds This is the time is takes for the glass to basically complete the latter half of an imaginary parabola. Now, I don't know if I can assume this part. R is the range, or the total x component of a parabola. If this trajectory where to be a parabola, R would equal twice of 1.40, which is 2.80. And the total time of would ideally be twice the time it took to get from the highest point to the ground (2* .419) and that is .838. Given R = vxi(total time), the initial x component of velocity (which, I was told is constant in the chapter I am covering) is 3.34 (roundabout). I think I'm stretching it a bit too much, and there is probably an easier way to doing that problem, but, well, that's what I thought of. If anyone would please tell me if my train of thought is correct, or, if it is incorrect, where did I go wrong? Can it be done this way? Why? Why not? (b) I'm not sure how to figure this part out. Nor do I exactly know what this question is asking for.