# (Mechanics) Getting a distance in a Ball Drop question!

1. Sep 13, 2011

### kid0

1. The problem statement, all variables and given/known data

A ball is dropped from the roof of a building. An observer looking outside the window sees the ball just outside. If the window is 3.99 m tall and the ball is in view for 0.897 s, how far below the roof is the bottom of the window?

2. Relevant equations

"Kinematics Equations
average velocity:
vave=Δx/Δt

average acceleration:
aave=Δv/Δt

Uniform Acceleration:

vave=(vi+vf)/2

Three key variables (displacement, time, velocity) lead to three key relationships relating each pair:

velocity & time:
v=v0+at

displacement & time:
x=x0+v0t+(1/2)at2

velocity & displacement:
v2=v20+2aΔx"
source : Physicsforums.com

3. The attempt at a solution

I do not know how to even start this question from..
For example, what equations to use or what numbers do I need to get in order to solve this question.

2. Sep 13, 2011

### SophusLies

If I get stuck in any physics problem, the first thing I do is make a list of knowns and unknowns. What are your knowns?

3. Sep 13, 2011

### kid0

mm.. knowns are..
I actually am confused with the "window is 3.99m tall".. if it means the window itself is 3.99 m tall OR it's up at the height of 3.99m from the ground.. if it means itself is 3.99m tall,
I think knowns would be the
h of window = 3.99
t of ball while in sight of the person = 0.897
V0 = 0 m/s
a = -9.80 m/s^2

??

4. Sep 13, 2011