# Finding time with distance, initial velocity and acceleration

1. Jan 16, 2017

### Justin Che

1. The problem statement, all variables and given/known data
A stone is dropped from the top of a tall building. It accelerates at a rate of 9.81m/s^2. How long will the stone take to pass a window that is 2.0m high, if the top of the window is 20.0m below the point from which the stone was dropped?

I've already solved the initial velocity when it reached to the window which is 19.81m/s.
So what i needed to solve now is the time when it pass the window that is 2m high.

vi=19.81m/s
d=2M
a=9.81m/s^2
t=?

2. Relevant equations
d=Vot+1/2at^2

3. The attempt at a solution
d=Vot+1/2at^2
2m=19.81m/s(t)+(0.5x9.81m/s^2)t^2
0.1=t+(4.9m/s2)t^2
0.02=t+t^2

The answer is 0.1s, but Im having trouble finding the right answer.

Last edited: Jan 16, 2017
2. Jan 16, 2017

### Staff: Mentor

Hi Justin Che. Welcome to Physics Forums.

You need to make an attempt before help can be given. Your relevant equation looks promising...