How are the hydrogens attached to the nitrogen atom in the ammonium ion?
The nitrogen atom, on both ammonium ion ($\ce{NH4+}$) and ammonia ($\ce{NH3}$), is $\ce{sp^3}$ hybridized. That's because we have 4 sigma bonds on ammonium ion and 3 sigma bonds + 1 electron pair in ammonia. The most stable orbital configuration for this set up is a tetrahedral shape.



Nitrogen can have a $\ce{sp^2}$ configuration if it has 3 sigma bonds and a pi bond, like in $\ce{H2C=NH2}$. The $\ce{sp^2}$ orbitals are "used" in the sigma bonds and the remainder $2p_z$ orbital is "used" in the pi bond. This gives a triangular plane shape to the molecule.



Finally, nitrogen can also have a $\ce{sp}$ configuration if it has 1 sigma bond and 2 pi bonds, in $\ce{N2}$ for example. In this case, the $\ce{sp}$ orbital is "used" in the sigma bond and the remainder $2p_y$ and $2p_z$ orbitals are "used" in the pi bonds. This gives $\ce{N2}$ a linear geometry.



Molecular orbital theory and hybridization are a little hard to process (at least for me it was and it's still being), so take your time to read some more about it, but at least with those examples you should have an idea of what to expect from some molecules geometries.