How Fast Will the Flower Pot Hit the Ground?

[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 L_w. 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, L_w, and g.

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

V_b=L_w/t+(g*t)/2.

Express your answer in terms of some or all of the variables h_b, L_w, t, v_b, and g.


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

h_b=v_b+1/2(gt²)
the equation given...
v_g=v_b+g*t
v_b=1/2*g*t²-h_b

I keep missing one variable... and I'm getting frustrated... :'(
Please help me out.

Thank You,
tomboi03

 

[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.

 

[nlightner]

I understand your frustration and I am here to help you out. Let's break down the problem and see if we can come up with a solution.

First, let's define our variables:
t = time the pot is visible
Lw = vertical length of your window
g = acceleration due to gravity
hb = height of the bottom of the window above the ground
vb = speed at the bottom of the window
vg = velocity of the pot as it hits the ground

We can use the equation hb = vb + 1/2(gt^2) to solve for vb. We know that at the bottom of the window, the velocity is 0, so we can rewrite the equation as:
hb = 0 + 1/2(gt^2)
Solving for vb, we get:
vb = (2hb/g)^0.5

Now, we can use the equation vg = vb + gt to solve for vg. Substituting our value for vb, we get:
vg = (2hb/g)^0.5 + gt

Since we know that the pot was dropped from the floor above, we can assume that the height of the bottom of the window (hb) is equal to the height of the floor above. So we can rewrite the equation as:
vg = (2(hb+Lw)/g)^0.5 + gt

Now, let's substitute the equation given in the problem for vb:
vb = Lw/t + (g*t)/2

Substituting this into our equation for vg, we get:
vg = (2(hb+Lw)/g)^0.5 + gt = (2(hb+Lw)/g)^0.5 + (Lw/t + (g*t)/2)t
Simplifying this equation, we get:
vg = (2(hb+Lw)/g)^0.5 + Lw + (g*t^2)/2

So, the velocity of the pot as it hits the ground (vg) can be expressed in terms of the variables hb, Lw, t, g. I hope this helps you in solving the problem. Good luck!