We can safely ignore the acceleration time, so the distance is just terminal velocity times time. Catching up means equal distance for both, so

v1*t1 = v2*t2

Substituting t1 with t2+T, where T is the headstart, we can solve for t2:

v1*(t2+T) = v2*t2
v1*T = v2*t2-v1*t2
t2 = (v1*T)/(v2-v1)

v1*T is also the distance of the first jumper when the serving jumper jumps. So it’s headstart distance divided by the relative velocity.

What’s curved, yellow, normed and complete?

A bananach space