1. Given data

• Mass of solid sphere: $m = 2\,\text{kg}$
• Total kinetic energy: $K = 2240\,\text{J}$
• The sphere is in pure rolling on a horizontal surface.

2. Kinetic energy of a rolling body

For pure rolling, $K = K_{\text{trans}} + K_{\text{rot}}$

So, $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a solid sphere, moment of inertia about its centre is $I = \frac{2}{5}mR^2$

Also, for pure rolling, $v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R}$

3. Substitute into the kinetic energy expression

$K = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2$

$K = \frac{1}{2}mv^2 + \frac{1}{2}\cdot \frac{2}{5}m v^2$

$K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$

$K = \left(\frac{5}{10} + \frac{2}{10}\right)mv^2$

$K = \frac{7}{10}mv^2$

4. Put the given values

$2240 = \frac{7}{10}\times 2 \times v^2$

$2240 = \frac{14}{10}v^2 = 1.4v^2$

$v^2 = \frac{2240}{1.4} = 1600$

$v = 40\,\text{m s}^{-1}$

5. Final answer

The velocity of the centre of mass is $\boxed{40\,\text{m s}^{-1}}$