# Falling past a window

1. Feb 4, 2009

### tomboi03

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

So the known variables are... t, Lw, and g.

If the bottom of your window is a height h above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb , the speed at the bottom of the window, defined by

Vb=Lw/t+(g*t)/2.

Express your answer in terms of some or all of the variables hb, Lw, t, vb, and g.

I have these equations already... but i don't know where to go about from here....

hb=vb+1/2(gt2)
the equation given....
vg=vb+g*t
vb=1/2*g*t2-hb

I keep missing one variable... and i'm getting frustrated... :'(

Thank You,
tomboi03
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2009

### Delphi51

Okay, so you know the speed at the bottom of the window, h above the ground. You just have to find the speed after it falls that distance h.

"hb=vb+1/2(gt2)" doesn't make sense - it says a distance equals a velocity.
"vg=vb+g*t" would be correct if t was the time to fall the height h, but it is not the same as the t in the question.
"vb=1/2*g*t2-hb" has a distance subtracted from a speed, which is impossible because the units clash.

What you really need here is an accelerated motion formula with initial and final velocities, distance, but not time. If you can locate one on your list, it will solve the question instantly. Failing that, you must use a distance formula like
d = Vot + .5at^2 to find the time to fall the height h. Then use a velocity formula like
V = Vo + at to find the velocity after that time.