Modeling bouncing using a Dirac delta function For a ball that is dropped in the presence of a gravitational field $mg$,
with a resistive term $-kv$, how can I find the equation of motion, given that when the ball hits the ground at $y=0$ it reflects with the same momentum instantaneously.
I have tried to model this force as  $-2mv\,\delta[y(t)]$
this function would return a $\delta(0)$ for values of $t$ that correspond to a $y=0$ *i.e. when the ball hits the ground), which would correspond to a force, and when $y(t)$ is not zero there would be no force as expected intuitively.
Originally I naively assumed that the integral $dt$ of this is $-2mv$ (impulse) which would correspond to an elastic reflection, but then I realized that $v$ is also a function of $t$, and thus the integration isn't as simple as saying $v$ is independent, and that the $\delta$-function's integral is 1.
Trying to do this integral the correct way  is $-2m\int v(t) \delta[x(t)-0]\, dt$.
Now normally the way you tackle this is if it were, e.g.,
$$\int v(t)\delta(t-t_{0})\,dt,$$
then obviously this integral would be $v(t_{0})$.
However this isn't a variable $t$, it's a function of $t$, so the thing I would need to insert into $v(t)$ is the value of $t$ that makes the $\delta$-function equal to zero, $x^{-1}(0)$.
So the impulse of this force would be
$$-2mv[x^{-1}(0)],$$
which intuitively makes sense, as $x^{-1}(0)$ represents the values of $t$ that correspond to the velocity of my ball when it hits the floor, and thus giving me the correct change of momentum in my model for when it strikes the ground.
However, $x^{-1}(0)$ is not a function and has many values of $t$, for when the ball then goes back up and down and so forth.
So my question is, is this model correct. and if so how can you then use it to solve my equation? And if not, what is the correct way to deal with this?
 A: The easiest way to do determine the equation of motion is probably to model the free motion of the ball separately from the ground collision. That is, solve for when the ball initially hits the ground, perform the elastic collision by inverting the velocity, solve for when the ball will hit the ground again, invert the velocity, and so forth. It is possible to write down what the collision force would be (see note below), but it's not particular useful in actually computing the motion itself. Since collisions are discrete events that cause a sudden jump in velocity, it inherently introduces discrete cases to consider if you attempt to setup and solve the differential equation.
Note about computing the force applied by the ground:
It's important to model the $t$ dependence inside the delta function carefully (e.g. $\int \delta(at) dt \neq \int \delta(t) dt$). Consider a small time $t_-$ before the collision and a small time $t_+$ afterwards so that we can neglect any other forces besides the collision force in this time interval. The momentum $p(t)$ in this time interval is given by $p(t) = mv (2\Theta(t - t_0) - 1)$ where $t_0$ is the collision time and $\Theta(t)$ is the step function. The collision force is thus given by $F(t) = dp/dt = 2mv \delta(t - t_0)$, which is not equivalent to $ 2mv\delta(x(t) - x_0)$.
