Okay, we're learning relative motion in class and I just want to see if I am uderstanding things correctly. Okay there is a guy in an elevator and the elevator is on the ground. Also on the ground (outside of the elvator) is a ball. A ball is shot upward with Vo = 7 m/s. At the same time, the elevator moved up from the ground at a constant speec of Vc = 3 m/s. I have to find the maximum height (Ym) the ball reaches relative to a) the ground and b) the floor of the elevator? Then I have to find the rate at which the speedo f the ball changes relative to c) the ground and d) the elevator floor? Well, this is what I am understanding: a) relative to the ground, I can ignore the speed of the elevator so: V^2 = Vo^2 + 2aYm 0 = 7 -19.6Ym Ym = 0.357m b) relative to the floor of the elevator cab, I am thinking that since the cab and the ball are both going up and the ball has a greater initial velocity: V^2 = 0 = (Vo - Vc)^2 - 19.6Ym 0 = 4^2 - 19.6Ym -16 = -19.6Ym Ym = 0.816m For c and d I am to find the rate at which the speed of the ball changes. Rate usually deals with time, right? And the change of speed over time is an acceleration, right? Then can I do the following: c) V(t) = Vot +0.5 at^2 0.357 = 7t -19.6t^2 t = 1.38s or t = 0.053s Then can I take the initial Vo and divide by those times? Vo/t1 = 7/1.38 = 5.07 m/s^2 Vo/t2 = 7/0.053 = 132 m/s^2 This sounds rather odd. Can someone point me the right direction? Thanks!