# Gravity as a Function of X

1. Dec 14, 2007

### NerfMonkey

This isn't a homework question, just one that's been nagging me for a while, and it is algebra so I thought this an appropriate place for it.

I've been working on this problem at my whiteboard for a good half hour and can't figure out. I'm sure it's very simple but a search of Google and these forums uncovered nothing of any value.

I want to find the point at which an object's velocity will reach 0 if its initial velocity is known and gravity acts at its normal $$-9.8m/s/s$$. The problem is that I don't know how to express this as $$f(x)$$ or $$y$$. I've read that gravity is $$9.8m/s^2$$ but I really don't understand how that works.

I apologize for asking such a basic question, I just don't remember anything from the last quarter of my algebra class. Can anyone help me out? Thanks in advance.

I went a little overboard with the tex thing. Oh well. $$:)$$

2. Dec 15, 2007

### ice109

how is this an algebra question? there are very simple formulas out there for this problem.

3. Dec 15, 2007

### HallsofIvy

Staff Emeritus
You are saying that the body's acceleration (derivative of velocity) is -g.
$$\frac{dv}{dt}= -g$$ so, integrating, v= -gt+ C. Taking the "initial speed" to be v0, v(t)= -gt+ v0. To determine when the objects velocity is 0, set that equal to 0 and solve for t: -gt+ v0= 0 so -gt= -v0 and t= v0/g.

Since this is neither abstract nor linear algebra, I am moving it.

Last edited: Dec 15, 2007
4. Dec 15, 2007

### Gib Z

My answer: when x=0, assuming the object is inelastic.

EDIT: Ahh nvm you meant on the objects way up, I thought you meant after it has already started falling.

5. Dec 15, 2007

### NerfMonkey

Thanks for the help.

I thought this was an algebra question, sorry it was in the wrong place. Shows how little I know about math.

6. Dec 15, 2007

### Feldoh

It essentially algebra, just not linear algebra/abstract algebra

7. Dec 15, 2007

### Rockyt

ok, I'm assuming you mean that you are shooting an object straight up with some initial velocity. I'd use conservation of energy to solve this.

at the point where v=0 all of our initial kinetic energy is converted into gravitational potential energy:
$$\frac{1}{2}mv^{2}=mgy$$

we solve for y:
$$y=\frac{v^{2}}{2g}$$

And there you have "the point at which an object's velocity will reach 0". It is important to note that this only works when gravity really is 9.8 m/s^2. If you're shooting things off into space then you must take into account that g is not constant so it gets a bit more complicated.