Rubber ball dropped from the roof of skyscraper reached terminal velocity v=75m/s?

Question

and elasctically bounced upward from concrete sidewalk.

What is velocity of the ball 1/4 second after the bounce?

g = 10m/s2

Answer 1

First answer disregards acceleration (af) due to air friction which immediately after the bounce = -10 m/s^2 (- = downward).
At any v going upward, af = -10*(v/vterm)^2 m/s^2, atotal = -10*(1+(v/vterm)^2) m/s^2
I can easily write the difference equations for a time-stepped simulation, but this is not explicitly integrable (at least by me).
An approximation, assuming constant acceleration = -2g, predicts the velocity to be 75 - 2gt = 70 m/s.

Answer 2

Ok, I'll say it.

72.5m/s