Velocity when touching ground and when fully stopping at arena So my doubt is: my teacher told us that when a ball hits the ground, the velocity isn't 0, but I've encountered a problem that involves a ball falling (free fall) to the sand, and when hitting the sand the ball receives a negative acceleration of a=-605m/s, and then stops fully. At this last problem, the teacher used the fact that when it stops (underground), the velocity is 0, while when from a free fall, the velocity when hitting the ground isn't 0. Why is this?
 A: Both cases are in fact more similar than you may think. Just before hitting the surface, the velocity of the ball is of course not zero. Then, the ball hits the surface, but it does not stop instantly. The precise description of what happens may be complicated depending on how precisely you want to model it. The most simple thing may be to say that the ball receives a negative acceleration from the surface, just as your teacher did in the case of sand, but you can extend this idea to other kind of surfaces. With a little bit of handwaving, one can say that a "hard" surface will give the ball a bigger acceleration than a "soft" one (like sand). From there you can calculate how deep the ball goes before being completely stopped; and you can wonder what happens next. Will it bounce for instance ? That will once again depend on your model.
A: Any real ball has some degree of elasticity, that is it is not 100% rigid. So, the moment that it hit the ground corresponds to the maximum falling velocity. From that moment, the ball starts to deform and decelerate, until stops and starts bouncing back. Here we consider that the ground is much more rigid than the ball.
But, if the ground is softer as in the case of sand, its the ground that (mostly) deforms until the ball stops.
Depending on the velocity of the ball that deformation can be huge, as shown by deep meteorite craters.
A: 
At this last problem, the teacher used the fact that when it stops
(underground), the velocity is 0, while when from a free fall, the
velocity when hitting the ground isn't 0. Why is this?

Because the ground has done work on the ball bringing it to a stop. According to the work energy theorem the net work done on an object equals its change in kinetic energy, or
$$W_{net}=F_{ave}d=\frac {mv_{f}^{2}}{2}-\frac {mv_{i}^{2}}{2}$$
$F_{ave}$ is the average force exerted by the ground as the ball penetrates the ground, $d$ is the distance the ball moves into the ground, $v_i$ is the initial velocity of the ball upon impact ($=mgh$), and $v_f$ is the final velocity of the ball equal to zero. Thus
$$F_{ave}d=-\frac {mv_{i}^{2}}{2}$$
The work is negative because the force exerted by the ground is opposite to the displacement $d$. This essentially means the work done by the ground has removed kinetic energy from the ball. This assumes the distance $d$ into the ground is much less than the height $h$ from which the ball fell so that we can neglect the loss of gravitational potential energy over the distance $d$.
Hope this helps.
