Saturday

August 2, 2014

August 2, 2014

Posted by **Physics** on Tuesday, June 23, 2009 at 4:20pm.

0 = Xo + 2^-1 a t^2

for t

I've been told that it is

(a^-1 2x)^(2^-1)

but I do not see how please show me step by step how you get that

0 - Xo = Xo - Xo + 2^-1 a t^2

(-Xo = 2^-1 a t^2) a^-1 2

(a^-1 (2-Xo) = t^2)^(2^-1)

(a^-1 (2 - Xo))^(2^-1)

please show me how to do this as i'm getting negetive number imaginary

show me step by step how to rearang please not just the rearanged equation but how you rearange it because i do not get it

- Physics -
**bobpursley**, Tuesday, June 23, 2009 at 4:23pmsee mathmates response to the question before. I dont understand how you make it so complicated.

x=x0+ vo*t +1/2 gt^2

let x0, vo be zero (so x downward will be negative).

x=1/2 g t^2

t=sqrt (2x/g)-8.8 sec about

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