Salmon often jump waterfalls to reach their breeding grounds.

Question

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.26 m away from a waterfall 0.509 m in height, at what minimum speed must a salmon jumping at an angle of 27.5 ◦ 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

Let

α = 27.5° be the jump angle,

v (to be calculated) be the initial speed,

g = 9.81 m/s² be gravitational acceleration,

t be time variable, starting with t = 0 at the moment of jump.

The salmon follows a parabola resulting from a combination of two movements:

- A straight line at angle α with constant speed v due to inertia,

which is what the salmon would follow in the absence of gravity.

- A vertical fall due to gravity, which is what the salmon would follow

in the absence of any initial velocity.

You can view the trajectory in an X-Y coordinate space, where

the x coordinate is horizontal, and

the y coordinate is vertical, with the origin at the point of jump.

Then [x(t), y(t)] is the position of the salmon after time t.

The horizontal distance x(t) is unaffected by the vertical fall, therefore

x(t) = v×t×cos(α)

In contrast, the vertical distance y(t) is a combination of the two movements:

y(t) = v×t×sin(α) - g×t²/2.

The portion "v×t×sin(α)" is the height the salmon would reach after time t

in the absence of gravity, and

"g×t²/2" is the amount of fall the salmon would suffer without any initial velocity.

The flight continues until the salmon hits the top of the waterfall at some time t1,

which is constraint by two requirements:

x(t1) = 2.26

y(t1) = 0.509

After substituting x(t) and y(t) we get the two equations

v×t1×cos(α) = 2.26

v×t1×sin(α) - g×t1²/2 = 0.509

From the first equation

t1 = 2.26/v×cos(α)

Substitute t1 into the second equation and simplify

2.26×tan(α) - g×2.26²/2v²×cos²(α) = 0.509

Express v

v = √(g×2.26²/2cos²(α)×(2.26×tan(α) - 0.509))

= (2.26/cos(α))×√(g/(2×(2.26×tan(α) - 0.509)))

Substitute actual numbers

v = (2.26/cos(27.5°))×√(9.81/(2×(2.26×tan(27.5°) - 0.509))) = 6.9 m/s