Question

Salmon often jump waterfalls to reach their

breeding grounds.
Starting 3.16 m from a waterfall 0.379 m
in height, at what minimum speed must a
salmon jumping at an angle of 37.6
◦
leave the
water to continue upstream? The acceleration
due to gravity is 9.81 m/s
2
.
Answer in units of m/s.

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 - (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=(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{(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`