If wave packets spread, why don't objects disappear? If you have an electron moving in empty space, it will be represented by a wave packet. But packets can spread over time, that is, their width increases, with it's uncertainty in position increasing. Now, if I throw a basketball, why doesn't the basketball's packet spread as well? Wouldn't that cause its uncertainty in position to increase so much to the point it disappears?
EDIT: I realize I wasn't clear what I meant by disappear. Basically, suppose the wave packet is spread over the entire Solar System. Your field of vision covers only an extremely tiny part of the Solar System. Therefore, the probability that you will find the basketball that you threw in your field of vision is very small.
 A: Short answer is it won't disappear because the integral of the probability density is still 1 even for a highly spread wavepacket, i.e. the object will still be found somewhere.
Slightly longer answer is that, if I start with a Gaussian wavepacket with width $a$, then after time $t$, the width will have spread to $$\sqrt{\frac{a^2+\hbar^2t^2/m^2}{a}}$$ The incredible smallness of $\hbar$ makes the spread negligble for something as massive as a basketball.
A: It is true that the spreading depends on the mass as @twistor59 has already noticed, but the more important fact is that the basketball is an open system and interaction with its surrounds makes that (due to decoherence) the state of the basketball is not described by quantum wavefunction theory [*]. Using the Wigner-Moyal formulation of quantum mechanics it is possible to show that the basketball always have a well-define position $x(t)$ at each instant.
[*] Wavefunction theory only applies to isolated quantum systems. 
A: Spreading the wave packet does not mean spreading and disappearing the electron.
If a basketball has initially uncertainty $\Delta V$ of its velocity, then with time the ball position uncertainty will grow as $\Delta V\cdot t$.. With time this uncertainty gets so large that the ball disappears from your sight (I mean you will not find it where you expect it to be without knowing the initial velocity spread).
A: AFAIK there is no answer to this question, that is why there are few "theories" that try to answer this question like GRW
GRW
EDIT:
Let me just elaborate because I gave you the most interesting answer without directly addressing your question. There is no need for you to throw the ball. The atoms of the ball are constrained by their mutual potential so the waves do not spread in the sense of free particles. @Twistor59 answer is heuristic  usually given in textbooks, but obviously there are no 100 g particles as such. The main issue is why we don’t see the ball at 100 meters or whatever since wavefunctions have non zero probability throughout space. The measurement problem is a close relation to your “proper “question but a bit different. GRW is more concerned with that. 
