# Kinematics Free fall problems part 2 help

1. May 10, 2010

### P944

1. The problem statement, all variables and given/known data
While riding on an elevator descending with a constant speed of 3.3 m/s, you accidentally drop a book from under your arm.

(a) How long does it take for the book to reach the elevator floor, 1.0 m below your arm?

(b) What is the book's speed relative to you when it hits the elevator floor?

2. Relevant equations
v = v0 + a*t
x = x0 + v0 *t + 1/2*a*t^2
x= x0 + v*t-1/2*a*t^2
V^2 = v0^2 +2*a*(x-x0)

3. The attempt at a solution

I tried to solve for time by first solving for final velocity using V^2 = v0^2+2*a*(x-x0) =
V^2 = (3.3m/s)^2 +2(-9.80m/s^2)(0-1m)
V^2 = 30.49 m^2/S^2
square root of 30.49 = 5.522 m/s

v = v0 +a*t
5.522 = 3.3 + (9.80)*t
t = 0.2267

part b) substract the book's velocity 5.522 - my velocity 3.3 = 2.222

I got both answers wrong. I was wondering if someone could help me explain what i did wrong? thank you so much!

2. May 10, 2010

### Char. Limit

Shouldn't your initial velocity be negative, considering that you're descending?

3. May 10, 2010

### Logistix

Since you're dropping the book with an initial velocity, and the rest of the falling path is accelerating, the velocity at any moment can be calculated, as you know, with:
v = v0 + at
This, however, is the velocity. If we want the distance, we have to multiply by t (vt=s)
v = v0 + at /*t
s = v0t + at^2/2
Now you can substitute s with the height of 1:
1 = 3.33t + 9.81t^2/2
Go on from here. If you have trouble extracting t which you need, it can be substituted:
v = 3.33 + 9.81t
v - 9.81t = 3.33
t = v-3.33/9.81
Insert it into the previous expression:
1 = 3.33(v-3.33/9.81) + 9.81*(v-3.33/9.81)^2/2
Hope you can work it out from here (you'll get the final v and from that the t).
Good luck!