Encompassing air resistance into free fall.?

Question

I've calculated the drag constant for a tennis ball (v^2 * -6.15x10^-4) and we know that vf=vi+at and df=1/2at^2+vit+di. But how would I be able to calculate the velocity and displacement at any given time, if I encompassed air resistance?
To anyone who wants to know, you can find the force of drag at any given time by: (vi+at)^2 or (-9.8t+vi)^2, where down is negative, and multiply that by the drag constant. So if you through a ball UP in the air at 30 m/s, the drag at any given time would be: f(x)=(-6.15x10^-4)*(-9.8t+30). So I want to know how to put this together in one function to find the displacement or velocity at any given point, including air resistance. Thanks!

Answer 1

There's no way to incorporate the nonlinear drag force into a closed-form equation. You'll need a (relatively simple) time-stepped simulation in which the acceleration is computed at each step based on current velocity, drag and gravity. It'll be something like the old gunnery sims that came out in the early days of computing. This is the general idea:
do
t = t+deltat
drag = f(previous v)
a = f(drag,g,etc.)
v = previous v + a*deltat
s = previous s + v*deltat
loop
Of course there are more sophisticated integration schemes than this which reduce errors considerably.

Answer 2

In a loose fall, the acceleration in that fall is a = g(a million - kV^2/W); the place g is g, ok is physique specific (e.g., coefficient of drag and go sectional area, plus air density), V is the linked fee of the air flowing over the physique (e.g., the linked fee of the autumn in nonetheless air), and W = mg is the load of the object with mass m. As you will see that, while kV^2 = W, a will become = 0 and the physique no longer quickens in its fall. while that happens we are saying the physique has reached terminal speed. And which would be discovered from V = sqrt(W/ok). for standard situations, a human physique reaches V ~ a hundred and twenty mph terminal speed while it falls from a extreme sufficient top. kV^2 = D is the air resistance, that's oftentimes called drag while speaking approximately aviation. As you will see that from a = g(a million - D/W), drag has no result on g the gravity field or the load or mass of the object. even with the undeniable fact that it does decelerate the acceleration so it is going to ultimately be 0 if allowed to fall far sufficient. And sure, that's a sort of friction stress.