1. Speed of the swing at the lowest point

When the swing is released from angular amplitude $\theta_0$, the person descends by height $h = L(1-\cos\theta_0).$ So at the lowest point, by energy conservation, $\frac12 Mv^2 = MgL(1-\cos\theta_0).$ For small $\theta_0$, $\cos\theta_0 \approx 1-\frac{\theta_0^2}{2},$ thus $v^2 \approx 2gL\left(\frac{\theta_0^2}{2}\right)=gL\theta_0^2.$

So, $v^2 \approx gL\theta_0^2.$

2. What happens when the person stands up at the lowest point?

At the lowest point, the velocity is horizontal. If the person stands up, his center of mass rises vertically by $\ell$.

Since the swing length is very large compared to this shift ($\ell \ll L$), the new center of mass is effectively closer to the pivot by $\ell$. Hence the radius of circular motion becomes approximately $L' = L-\ell.$

At that instant, there is essentially no external torque about the pivot, so angular momentum about the pivot is conserved: $MvL = Mv'(L-\ell).$ Thus, $v' = v\frac{L}{L-\ell} \approx v\left(1+\frac{\ell}{L}\right).$

Hence, $v'^2 \approx v^2\left(1+\frac{2\ell}{L}\right).$

3. Increase in kinetic energy

Initial kinetic energy at the lowest point: $K_i = \frac12 Mv^2.$ Final kinetic energy after standing: $K_f = \frac12 Mv'^2.$ So,