Probably an easy problem but i'm having trouble- gravity

[tgoot84]

I think this will be easy for someone who understands this stuff:

While riding on an elevato descending with a constant speed of 2.6 m/s you accidently drop a book from under your arm.

a. How long does it take for the book to reach the elevator floor, 1.1m below your arm?
(in seconds)

b. What is the book's speed (relative to the earth) when it hits the elevator floor?
(m/s)

I appreciate the help

 

[chroot]

Since the elevator is descending at a constant velocity, the physics inside it is indistinguishable from "normal" physics experienced when at rest. In other words, the book will fall in the same duration as it would if the elevator were not moving at all. The book's velocity, with respect to the earth, is just the book's velocity with respect to the elevator + 2.6 m/s.

- Warren

 

[tgoot84]

great that makes much more sense, thanks

 

[HallsofIvy]

Originally posted by chroot
Since the elevator is descending at a constant velocity, the physics inside it is indistinguishable from "normal" physics experienced when at rest. In other words, the book will fall in the same duration as it would if the elevator were not moving at all. The book's velocity, with respect to the earth, is just the book's velocity with respect to the elevator + 2.6 m/s.

- Warren

Yes, "it is indistinguishable from "normal" physics experienced when at rest" but why would the book's velocity with respect to the elevator, when it hits the floor of the elevator, be 2.6 m/s? That was given as the speed of the elevator with respect to the Earth and has nothing to do with the speed of the book relative to the elevator.

There are two ways to do this problem:
1) The EASY way: the book, falling, experiences acceleration -9.8 m/s². It is initially not moving relative to the elevator so after t seconds it will have fallen -4.9t² m and will have velocity -9.8t m/s(both relative to the elevator). Since it has to fall 1.1 m to hit the floor of the elevator, the time is given by -4.9t²= -1.1 or t= 0.474 seconds. In that time it will have accelerated to -9.8(0.474)= -4.6 m/s (relative to the elevator).

2) The HARD way: the book, falling experiences acceleration -9.8 m/s². It is initially moving with velocity -2.6m/s relative to the earth. After t seconds, it will have velocity -9.8t- 2.6 m/s and will have gone a distance -4.9t²- 2.6t (both relative to the earth). In t seconds, the elevator will have gone a distance -2.6t m/s so, in order for the book to hit the floor of the elevator, the book will have to have gone -2.6t- 1.1 meters. That is, we must have -4.9t²- 2.6t= -2.6t- 1.1. The "-2.6t" terms cancel leaving us with the same equation as before. That's (Gallilean) relativity for you! The book hits the floor of the elevator in 0.474 seconds as before.
The books speed, relative to the Earth is then -9.8(.474)- 2.6=
-7.2 m/s, relative to the earth, which is, of course, -7.2-(-2.6)= -4.6 m/s relative to the elevator.

 

[chroot]

Originally posted by HallsofIvy
Yes, "it is indistinguishable from "normal" physics experienced when at rest" but why would the book's velocity with respect to the elevator, when it hits the floor of the elevator, be 2.6 m/s? That was given as the speed of the elevator with respect to the Earth and has nothing to do with the speed of the book relative to the elevator.

That's not what I said. I said "The book's velocity, with respect to the earth, is just the book's velocity with respect to the elevator + 2.6 m/s."

- Warren

 

[HallsofIvy]

Originally posted by chroot
That's not what I said. I said "The book's velocity, with respect to the earth, is just the book's velocity with respect to the elevator + 2.6 m/s."

- Warren

Mea culpa, mea culpa. I interpreted it as "book's velocity with respect to the elevator: + 2.6 m/s." which on further consideration, was not Warren-ted.