The problem corresponds to the motion of a projectile (the salmon), with initial speed [tex]v_0[/tex], initial direction [tex]\theta=37.6^{\circ}[/tex] and vertical acceleration [tex]g=9.81 m/s^2[/tex] downward. The two equations which gives the horizontal and vertical position of the salmon at time t are
[tex]S_x (t) = v_0 cos \theta t[/tex] (1)
[tex]S_y (t) = v_0 sin \theta t - \frac{1}{2}gt^2[/tex] (2)
We can solve the problem by requiring Sx=3.16 m and Sy=0.379 m, the data of the problem.
Solving eq.(1) for t:
[tex]t=\frac{S_x}{v_0 cos \theta}[/tex]
And substituting this expression of t into eq.(2), we get the following expression for [tex]v_0[/tex]:
[tex]v_0 =\sqrt{\frac{g S_x^2}{2(tan \theta S_x -S_y)cos^2 \alpha}}[/tex]
And substituting the numbers into the equation, we find
[tex]v_0 = 6.16 m/s[/tex]
