Well, I was hanging out with a friend the other day in a shopping mall's foodcourt and this guy I know from school comes up to me, asks me if I have a pen and a paper, and writes out a physics question for me to solve (talk about being a weirdo ) and it went something like this: A man is juggling while riding on a horizontal escalator; he is moving at 1m/s relative to the escalator. The escalator is moving 0.5m/s relative to the ground. The maximum vertical point a ball reaches is 2m and another man, walking by the escalator, notices that the horizontal range of the ball is 1m. What is the velocity of the other man? (Relative to the ground) I guessed I'd first write down what's given relative to the ground: v ball x = 1.5m/s y = 2m xP = 1m Then, finding out how much time it takes for that one ball to go through half a cycle (i.e. in order to find ACTUAL horizontal range): 0 = vy^2 - 19.6*2 v ball y = 6.26m/s 0 = 6.26 - 4.9 t t = 1.27775s And to find the actual horizontal range, x = 1.5 * 1.27775 = 1.9166m And then, because everything in the x direction is going at constant velocity (and we're assuming so is the other man), so velocity is directly proportional to distance, and because the constant k, 1/t, is constant (because the time intervals are the same), we can write: k xP / kx = vP / v ball x = xP / x v P = 0.7826m/s Where v P is perceived velocity by the walking man and x P is the perceived range. Next, because he sees it as 0.7826m/s, we can write: V + 0.7826 = 1.5 V = 0.717m/s And what I got is that he's walking at 0.717m/s. Can anyone tell me if I'm right or wrong? Thanks.