A block of mass 5.0 kg is suspended from the end of a vertical...

At the equilibrium position, the spring force balances the weight of the block.

$k x_0 = mg$

$k (0.10) = 5.0 \times 10$

$k = 500 \text{ N/m}$

When the block is given an upward speed at the equilibrium position, it executes simple harmonic motion (SHM). The velocity at the equilibrium position is the maximum velocity of the SHM.

The angular frequency of the SHM is given by:

$\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{500}{5.0}} = 10 \text{ rad/s}$

The maximum height the block rises from the equilibrium position is equal to the amplitude $A$ of the SHM.

Using the relation v_{\text\max} = A \omega:

A = \dfrac{v_{\text\max}}{\omega} = \dfrac{2.0}{10} = 0.20 \text{ m} = 20 \text{ cm}