# Homework Help: Isolating t in horizontal motion equation

1. Jun 17, 2011

### mom2maxncoop

Isolating "t" in horizontal motion equation

1. The problem statement, all variables and given/known data
So this is just a practice question from my course and they gave me the answer, the problem I am having is understanding how they got to that answer. I have the formula and everything, just don't know how to isolate "t" and the steps to do so

-How long does it take for a flowerpot to fall from a balcony that is 18m above the ground?

2. Relevant equations

d=½a$\nabla$t²

3. The attempt at a solution

18m [down] = ½(9.8m/s²)$\nabla$t²
and I know I have to isolate the "t" but how do I go about doing that, I keep getting it wrong

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

2. Relevant equations

3. The attempt at a solution

2. Jun 17, 2011

### SteamKing

Staff Emeritus
Re: Isolating "t" in horizontal motion equation

Show us what you have done so far.

3. Jun 18, 2011

### mom2maxncoop

Re: Isolating "t" in horizontal motion equation

After posting this, I sat for another 2 hrs trying to figure out the problem and I finally understood it.

18m [down] = ½(9.8m/s²)∇t²
18(2)=9.8(t²)
36=9.8(t²)
36/9.8 = t²
3.67=t²
squareroot 3.67 = t
1.91=t