- #1

missrikku

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!