Question

Salmon often jump waterfalls to reach their

Answer 1

The problem corresponds to the motion of a projectile (the salmon), with initial speed $v_0$, initial direction $\theta=37.6^{\circ}$ and vertical acceleration $g=9.81 m/s^2$ downward. The two equations which gives the horizontal and vertical position of the salmon at time t are

$S_x (t) = v_0 cos \theta t$ (1)

$S_y (t) = v_0 sin \theta t - \frac{1}{2}gt^2$ (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:

$t=\frac{S_x}{v_0 cos \theta}$

And substituting this expression of t into eq.(2), we get the following expression for $v_0$:

$v_0 =\sqrt{\frac{g S_x^2}{2(tan \theta S_x -S_y)cos^2 \alpha}}$

And substituting the numbers into the equation, we find

$v_0 = 6.16 m/s$