Overshoot and peak time for 5/[s(s+4)]

Find the closed-loop transfer function, read off $\omega_n$ and $\zeta$, then use the standard second order formulas.

Closed-loop transfer function

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{5}{s^2 + 4s + 5}$$

Parameters

$$\begin{aligned} \omega_n^2 &= 5 \;\Rightarrow\; \omega_n = \sqrt{5} = 2.236\ \text{rad/s} \\ 2\zeta\omega_n &= 4 \;\Rightarrow\; \zeta = \frac{4}{2(2.236)} = 0.894 \end{aligned}$$

Since $\zeta < 1$, the response is underdamped.

$$\omega_d = \omega_n\sqrt{1 - \zeta^2} = 2.236\sqrt{1 - 0.8} = 2.236(0.4472) = 1\ \text{rad/s}$$

(The closed-loop poles are $s = -2 \pm j1$.)

Percentage overshoot

$$\%M_p = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 100\,e^{-\pi(0.894)/0.447} = 100\,e^{-2\pi} = 0.187\%$$

Peak time

$$t_p = \frac{\pi}{\omega_d} = \frac{\pi}{1} = 3.14\ \text{s}$$

Answer: $\%M_p \approx 0.19\%$ and $t_p = 3.14$ s. The system is heavily damped, so the overshoot is very small.