I'm not a math teacher, but I'm pretty good at figgering stuff out. I think you can pretty closely approximate the time it would take the ball to l30 feet given your parameters. I'll bet the approximization is pretty darn close.
Here's the formula:
Since T=D/V (time to travel equals the distance divided by the velocity), all we have to do is figure out the average velocity of the ball over the 130 feet.
By your formula, since the ball slows down by 1 MPH each ten feet, the average velocity in MPH would be the initial velocity minus half of the following: the distance (130 feet) divided by 10 (since the ball slows down one MPH each ten feet).
In formula terms: V(ave) = V(init.) - (D/10)/2
So, if we first compute average velocity in MPH and then convert it to feet per second.
V = 75 - ((130/10)/2))
Which equals 68.5 MPH average velocity.
68.5 ave velocity is almost exactly 100 feet per second (100.466 to be exact, which is 68.5 times 5280 divided by 3600)
So the time equals:
T = 130/100.466
or, just under 1.3 seconds
In terms of pop time, with a release of .7 or better, an arm strength of 75MPH would give us the sub 2.0 pop. Which makes sense.