Question

A physics student drops a marble (from rest) from some distance d above the top of a window. The freely falling marble takes t= 0.220 s to travel past a window h= 2.20 m tall. Use g=100 m/s2.

Answer 1

i think you can use \[\Delta y=v_{0}t+\frac{1}{2}at^2\]

Answer 2

First of all, I think they meant g=10.0m/s^2 instead of 100m/s^2
Then, I think that they forgot the question.

Answer 3

that's what i was thinking too. i just a statement as it is now. so i just said for them to use a formula that has only one variable missing with the given material. it might be asking for the height of the building. i don't know.

Answer 4

Yeah! you're right! It might be asking for the distance d above the top of the window(or the height of the building as you said)
So, if we assume that they are asking for the distance d. your equation is pretty good, except that they have to use a=g=-10.00m/s^2 as the sign convention of g is negative for a motion whose direction is downward.
Vo =0m/s (from the rest)
t=0.220s
Δ Y= Y-Yo, where Y=h=2.20m, and Yo=d, the value to be found.
The answer will be Yo=2.4m

Answer 5

Sorry! I meant:Delta Y=Y-Yo (The delta symbol is missing)

Answer 6

let's use g=-100. lol

Answer 7

Lol. I've never met a such high value for g, the acceleration due to the gravity close to the Surface of Earth.lolll...

Answer 8

haha it's so absurd it's funny.

Answer 9

Yeah! It is!