Question

Teaching myself physics... Can someone help haha? Im stumped on this one. show work please. all steps and thoughts


44.)A falling stone takes .28s to travel past a window 2.2m tall. From what height above the top of the window did the stone fall?

Answer 1

two equations for two unknowns.
once you find Vi, (and "initial" here refers to the velocity the stone has already when it first reaches the top of the window) you can use that to determine how far the stone has fallen up to that point...

you can use one of the equations we just used... remember that the velocity we just found is now going to be called "Vf" because we're considering the part of the trip from the top of the building to the top of the window
Vf^2 = (Vi^2 =0) +2g*h
find h

(changed the '-' to a '+' for consistency)

Answer 2

thanks! if I have anymore questions i'll post them here later :)

Answer 3

sure!

Answer 4

Nothing to add here, just happy to see another autodidact. :)

Answer 5

I changed a few of the signs in the equations in the sketch, it'll make it easier to solve because you won't have to be worrying about direction for velocity which can be a little confusing until you get used to it...

Answer 6

Vf^2 = (Vi^2 =0) +2g*h i dont understand the (Vi^2=0) part where is the =0 coming in?

Answer 7

also for my Vf in the Vf^2 = (Vi^2 =0) +2g*h is it 7.9? (gotten from dividing 2.2 and .28) and if so what is my Vi?

Answer 8

We have time and distance, so we can use \[s=v_{i}t+\frac{1}{2}at^2\] for\[2.2m = v_{i}(0.28s)+ \frac{1}{2}(9.8\frac{m}{s^2})(0.28s)^2\]. Now you have the initial velocity for when it got to the op of the window, which is the *final* velocity for the distance traveled from the drop to the top of the window. So, \[v_{f}^2= (0v_{i})^2+2as\]Substitute in the initial velocity from the first step for the final velocity in this one, and solve for the displacement. That should get you there.

Answer 9

@eighthourlunch actually did it in an easier way, I think. go with his method

Answer 10

thank you both!