# Rearranging the Projectile Motion Base Quadratic Formula for Initial Velocity

1. May 23, 2010

### Pearlhammer

1. The problem statement, all variables and given/known data
What I'm supposed to do is to rearrange the this formula -16t^2+Vt+h and solve it for V.
V= Initial Velocity t= time (throwing the ball in a parabolic arc) h= height
I know what the height is and it is 6ft. I also have the time which is 2.03 seconds.
How do I rearrange for V? I got an answer but I'm not sure and doubting if it is correct.
2. Relevant equations

-16t^2+Vt+h

-16t^2+Vt+6 (if you plug in the height.)

3. The attempt at a solution

-16t^2+Vt+6 What I started with

-16t^2+Vt= -6 Subtracted 6 to the other side

Vt= -6+16t^2 Added -16t^2 to both sides.

V= (-6/t)+16t Divided both sides (every term) by t. This is my answer so far.

I have no idea if I'm correct or not and I have the feeling I'm not. Please help, i'm in US grade 9.

Last edited: May 23, 2010
2. May 23, 2010

### Mentallic

Is that meant to be the physics equation for motion:

$$h=Vt+1/2gt^2$$ ?

So then re-arranging you have $$-1/2gt^2-Vt+h=0$$

It doesn't seem to be consistent with what you started with. But if we assume you started with teh correct equation, yes, you've done it right.

3. May 23, 2010

### Pearlhammer

Thank you. What I was talking about was just simply if I had the correct answer for:
Taking -16t^2+Vt+h
and rearranging for V to be by itself on one side of the equation.

4. May 24, 2010

### ktgster

-16t^2+Vt+h = 0
-16t^2+Vt = -h
Vt = -h + 16t^2
V = (-h + 16t^2)/t

your right op, but my way is better

5. May 24, 2010

### Pearlhammer

Thank you very much guys!

6. May 24, 2010

### Mentallic

How is your way better? It's exactly the same...