Well you would still include the m, but they will eventually cancel out, so omitting the m right from the start will not make a difference (although it is probably a bad habit to cancel things out before it actually happens during the process of the calculation).
What i mean is that initially, as the ball is released, it has kinetic and potential energy (with reference to the ground) given by:
[tex]E_k = 0.5 mv_1^2[/tex]
[tex]E_p = mgh[/tex]
And as it hits the ground:
[tex]
E_k = 0.5mv_2^2[/tex]
[tex]E_p = 0[/tex]
Once you create an equation from these terms, you will see that every term has an 'm' in it which can be eliminated by dividing everything by m.