Kinetic energy per unit of mass is
[tex]K=\frac{v^{2} }{2}[/tex]
Given, [tex]v=3m/s^{2}[/tex]
Therefore,
[tex]K=\frac{(3^{2} m/s^{2} )^2}{2}[/tex]
[tex]K=4.5 J/kg[/tex]
Now potential energy per unit mass is
[tex]p=g\times h[/tex]
Given, [tex]h=90 m[/tex]
Therefore,
[tex]p= 9.8m/s^2 \times 90[/tex]
[tex]p=882.9 J/kg[/tex]
Thus, total mechanical energy of the river water per unit mass is
[tex]T=K+p=(4.5+882.9)J/kg[/tex]
[tex]T=887.9 J/kg[/tex]
OR
[tex]T=0.887 kJ/kg[/tex]
