Answer:
[tex]v=10.34\frac{m}{s}\\\theta=73.14^\circ[/tex]
Explanation:
The velocity of the fish relative to the water when it hits the water is given by its magnitude and its angle:
[tex]v=\sqrt{v_x^2+v_y^2}\\\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]
The horizontal speed ([tex]v_x[/tex]) of the fish is constant, while vertically we have a free fall ([tex]v_{0y}=0[/tex]), using the kinematic equations we can calculate the vertical speed ([tex]v_y[/tex]):
[tex]v_y^2=v_{0y}^2+2gh\\v_y^2=2gh[/tex]
Now, we calculate the velocity of the fish:
[tex]v=\sqrt{v_x^2+2gh}\\v=\sqrt{(3\frac{m}{s})^2+2(9.8\frac{m}{s^2})(5m)}\\v=10.34\frac{m}{s}\\\theta=tan^{-1}(\frac{\sqrt{2gh}}{v_x})\\\theta=tan^{-1}(\frac{\sqrt{2(9.8\frac{m}{s^2})(5m)}}{3\frac{m}{s}})\\\theta=73.14^\circ[/tex]
