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I remember doing this in class, but forgot which equation to use

hmm could it be \[V_2 ^2 = V_1 + 2as\]
where s is the distance

correction.. \[V_2 ^2 = V_1^2 + 2as\]

V is velocity?
a is acceleration?
s is displacement?

yup

Oh... yeah so that's the same one I have here as
v^2=u^2+2as

yep same shiz

so first find the acceleration using
v = 0
u = v
s = X

0=v^2+2aX... what is a?

you solve for it

isolate a

a=-v^2/2X

right

now use
v^2 = u^2 + 2as again
this time use:
v = 0
u = 3v
a = -v^2/2X

then solve for s

starting with 0=9v^2+2aX
-9v^2/2a=X
-9v^2/2(-v^2/2X)=X
9X=X
now what?

no...

it's not 2aX

it's 2as

X is just for the first situation

oh i see

Oh. THank You!

welcome