# Relative motion of an elevator question

missrikku
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!

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HallsofIvy
Homework Helper
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
Your formula is correct (It took me a moment to recognize it. It's from "conservation of energy) but since v0 is 7, you should have
0= 49- 19.6Ym, not "7".

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
Yes, this time you remembered to square!

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?
Yes, the (c) asks for the acceleration relative to the ground which you have ALREADY used. It is -g, the acceleration due to gravity.

In (d), since the elevator has no acceleration, the acceleration of the ball, relative to the elevator, should be the same.
You can calculate it directly by using the velocity formulas:
for the ball, v(t)= 7- gt. For the elevator, v(t)= 4.
The velocity of the ball, relative to the elevator, is
v(t)= (7- gt)- 4= 3- gt. Now find the acceleration, either by differentiating: v'= -g, or, since the acceleration is a constant by calculating the average acceleration over any time:
when t= 0, v= 4. When t= 3, v= 4- 3g. The change in speed is
4-3g- 4= -3g and, since this happened in 3 sec., the average acceleration is -3g/3= -g (of course).