[tex]\text{Kinetic energy,}\\\\~~~~E_k = \dfrac 12 mv^2\\\\\implies v^2 = \dfrac{2 E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}m}\\\\\implies v = \sqrt{\dfrac{2 \times 12000}{40}}\\\\\implies v = 24.49~~ \text{ms}^{-1}\\\\\text{Velocity of the platform diver is 24.49 m/s}[/tex]
