Given,
The mass of the diver, m=51 kg
The height, 10 m
Let us assume that the initial velocity of the diver is u=0 m/s
From the equation of motion,
[tex]v^2=u^2+2gh[/tex]Where v is the velocity of the diver just before she hits the water and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} v^2=0+2\times9.8\times10 \\ \Rightarrow v=\sqrt[]{196} \\ =14\text{ m/s} \end{gathered}[/tex]The kinetic energy of the diver is given by,
[tex]E=\frac{1}{2}mv^2[/tex]On substituting the known values,
[tex]\begin{gathered} E=\frac{1}{2}\times51\times14^2 \\ =4998\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the diver just before she hits the water is 4998 J
