First, find the velocity of clay before impact using conservation of energy: $\frac{1}{2}m{v}^2=mgh$, so $v=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10\text{ m/s}$ downward.

During collision, momentum is conserved: $m_{\text{clay}}v=(m_{\text{clay}}+m_{\text{pan}})v_{\text{final}}$, so $0.5\times 10=\left(2.5\right)v_{\text{final}}$, giving $v_{\text{final}}=2\text{ m/s}$ downward.

Initial spring compression: $F_{\text{spring}}=k{x}_0=m_{\text{pan}}g$, so $200x_0=2\times 10$, $x_0=0.1\text{ m}$.

Now apply energy conservation after collision. Let maximum descent be $d$ from initial position. The change in spring potential energy is $\frac{1}{2}k{(x_0+d)}^2-\frac{1}{2}k{x}_0^2$, change in gravitational potential is $-(m_{\text{clay}}+m_{\text{pan}})gd$, and initial kinetic energy is $\frac{1}{2}(m_{\text{clay}}+m_{\text{pan}})v_{\text{final}}^2$. Set total energy change to zero:

$\frac{1}{2}k{(x_0+d)}^2-\frac{1}{2}k{x}_0^2-(m_{\text{clay}}+m_{\text{pan}})gd+\frac{1}{2}(m_{\text{clay}}+m_{\text{pan}})v_{\text{final}}^2=0$

Substitute values: $\frac{1}{2}\times 200\times {(0.1+d)}^2-\frac{1}{2}\times 200\times {0.1}^2-2.5\times 10d+\frac{1}{2}\times 2.5\times 2^2=0$

Simplify to $100{(d+0.1)}^2-1-25d+5=0$, then $100d^2+20d+1-1-25d+5=0$, so $100d^2-5d+5=0$.

Solve quadratic: $d=\frac{5\pm \sqrt{25-2000}}{200}$. Discard negative root, $d=\frac{5+\sqrt{-1975}}{200}$ not real? Correction: $100d^2-5d+5=0$ has discriminant $25-2000=-1975$, error in sign. Actually, from earlier: $100d^2+20d+1-1-25d+5=100d^2-5d+5=0$ is correct, but discriminant negative indicates math error. Re-check energy equation: initial compression energy is already stored, so term $\frac{1}{2}k{x}_0^2$ should not be subtracted? Actually, it is correct. The error is in constant: +1 -1 cancel, so $100d^2-5d+5=0$ gives $d=\frac{5\pm \sqrt{25-2000}}{200}$, but discriminant negative. The correct approach is to set the equation as $\frac{1}{2}k{(x_0+d)}^2=\frac{1}{2}k{x}_0^2+(m_{\text{clay}}+m_{\text{pan}})gd-\frac{1}{2}(m_{\text{clay}}+m_{\text{pan}})v_{\text{final}}^2$, but careful with signs. Actually, the energy conservation is: initial kinetic + initial spring potential + initial gravitational potential = final spring potential + final gravitational potential (with lowest point as reference). At initial position after collision, spring energy is $\frac{1}{2}k{x}_0^2$, gravitational potential is 0 (reference), kinetic is $\frac{1}{2}\left(2.5\right)2^2=5\text{ J}$. At lowest point, spring energy is $\frac{1}{2}k{(x_0+d)}^2$, gravitational potential is $-\left(2.5\right)gd$, kinetic is 0. So:

$\frac{1}{2}k{x}_0^2+5=\frac{1}{2}k{(x_0+d)}^2-25d$

Substitute: $\frac{1}{2}\times 200\times 0.01+5=\frac{1}{2}\times 200\times {(d+0.1)}^2-25d$

$1+5=100{(d+0.1)}^2-25d$

$6=100d^2+20d+1-25d$

$6=100d^2-5d+1$

$100d^2-5d-5=0$

Solve: $d=\frac{5\pm \sqrt{25+2000}}{200}=\frac{5\pm \sqrt{2025}}{200}=\frac{5\pm 45}{200}$

Take positive root: $d=\frac{50}{200}=0.25\text{ m}$.

Final answer: 0.25 m