Say we have a ball that is dropped from a point $(a,h) $ and falls under acceleration $g$ onto a function $f(x)$, where it hits the function at the point $(a,f(a))$, $h>f(a)$.
It will then follow a parabolic path that is defined by the gradient at the point it hits the function and the height above the function in which it is dropped. What is the equation of this parabola in the form $y=Ax^2+Bx^2+C$?
Here is a picture of what I am trying to explain: (In this scenario, $f(x)=sinx$ )
I made a desmos graph that allows you to see the trajectory of a ball dropped from a point above any function $f(x)$ that the user may input. Here it is: https://www.desmos.com/calculator/9i65fwpllq
Using the law of reflection (angle of incidence = angle of reflection) and a bit of simple trig, you can find the angle '$\theta$' (to the horizontal) at which the projectile bounces off the function to be:
$$\theta=\tan^{-1}\left(\frac{f'(a)^2-1}{2f'(a)}\right)$$
We can also pretty easily find the velocity $v$ at which it bounces off the function using its height $h-f(a)$ above the function: $$v=\sqrt[]{2g(h-f(a))}$$
We can then sub our values into the parabolic equation of trajectory formula:
$$y=(x-a)\tan{\theta}-\frac{g(x-a)^2}{2v^2\cos^2{\theta}}+f(a)$$
And after simplifying, our final parabola becomes:
$$y= \left(\frac{(f'(a)^2+1)^2}{16f'(a)^2(f(a)-h)}\right)x^2+\left(\frac{f'(a)^2-1}{2f'(a)}-\frac{a(f'(a)^2+1)^2}{8f'(a)^2(f(a)-h)}\right)x+\left(\frac{a^2(f'(a)^2+1)^2}{16f'(a)^2(f(a)-h)}-\frac{a(f'(a)^2-1)}{2f'(a)}+f(a)\right)$$