Answer:

[tex]y_{max} = 0.829\,m[/tex]

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

[tex]v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}[/tex]

[tex]v = 2.913\,\frac{m}{s}[/tex]

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

[tex](3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}[/tex]

After some algebraic handling, a second-order polynomial is formed:

[tex]12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} – (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s[/tex]

[tex]1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s – 12.728 = 0[/tex]

The roots of the polynomial are, respectively:

[tex]\Delta s_{1} \approx 0.128\,m[/tex]

[tex]\Delta s_{2} \approx -0.099\,m[/tex]

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

[tex]\Delta s \approx 0.128\,m[/tex]

The maximum height that the block reaches after rebound is:

[tex](3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}[/tex]

[tex]y_{max} = 0.829\,m[/tex]