1. Sep 17, 2011

### SaltyBriefs

1. The problem statement, all variables and given/known data
I'm trying to derive this formula, but I get stuck after I factor the t out.

V$_{f}$$^{2}$ = V$_{0}$$^{2}$ + 2a (y-y$_{0}$)

2. Relevant equations

V$_{f}$$^{2}$ = V$_{0}$$^{2}$ + 2a (y-y$_{0}$)

3. The attempt at a solution
1) y$_{f}$ - y$_{0}$ = ($\frac{V_{0}+V_{f}}{2}$)t

2) y$_{f}$ - y$_{0}$ ($\frac{1}{t}$)= ($\frac{V_{0}+V_{f}}{2}$)t ($\frac{1}{t}$)

3) V$_{f}$= ($\frac{V_{0}+V_{f}}{2}$)t ($\frac{1}{t}$)

4) ????

5) V$_{y}$ = V$_{0y}$$^{2}$ + 2a (y$_{f}$-y$_{0}$)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 17, 2011

### PeterO

Firstly, I can only show this using symbols
v for your Vf - final velocity
u for your Vo - initial velocity
s for our y - yo - displacement.
{partially from familiarity, and partly because it is easier to type}

So I am aiming at v2 = u2 + 2as

you are hopefully familiar with a couple of other motion equations

v = u +at & s = t*(v + u)/2 [you listed tis second one in line 1 of your solution]

These two are combined.

The first can be transposed to give

t = (v - u)/a

substitute for t in the second

s = (v-u)(v+u)/2a

so

(v-u)(v+u) = 2as

v2 - u2 = 2as

or

v2 = u2 + 2as

3. Sep 17, 2011

### WatermelonPig

Alternatively:

a = dv/dt

a(ds) = ds/dt(dv)