Measuring the Collision Time!

Let $\langle v_{1,2}\rangle$ be the mean velocities during compression and decompression respectively. On the first interval, the speed changes uniformly from $\sqrt{2gh}$ to $0$ , and on the second interval, from $0$ to $\sqrt{2gh'}$ : $\vec{F}_{\text{compression}}=\text{constant}\implies \langle v_1\rangle=\frac{1}{2}\sqrt{2gh}$ $\vec{F}_{\text{decompression}}=\text{constant}\implies \langle v_2\rangle=\frac{1}{2}\sqrt{2gh'}$ Let the height of the spherical cap "flattened" on the ground at the time of maximum compression be $d$ . Some simple geometry yields: $\left(\dfrac{\psi}{2}-d\right)^2+\left(\dfrac{\varphi}{2}\right)^2=\left(\dfrac{\psi}{2}\right)^2$ Proceeding: $\tau_{1,2}=\dfrac{d}{\langle v_{1,2}\rangle}$ And therefore: $\tau_1=\dfrac{\frac{1}{2}\psi-\sqrt{\frac{1}{4}\psi^2-\frac{1}{4}\varphi^2}}{\langle v_1\rangle}=\dfrac{\psi-\sqrt{\psi^2-\varphi^2}}{\sqrt{2gh}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \tau_2=\dfrac{\psi-\sqrt{\psi^2-\varphi^2}}{\sqrt{2gh'}}$ And thus the total time of the collision can be expressed as: $\tau=\tau_1+\tau_2=\dfrac{\psi-\sqrt{\psi^2-\varphi^2}}{\sqrt{2g}}\left(h^{-\frac{1}{2}}+h'^{-\frac{1}{2}}\right)$ Numerically: $\tau\approx 872.966\:\mu s\approx \boxed{873\:\mu s}$