Starting downstream, 2.73 m away from a waterfall 0.504 m in height, at what minimum speed must a salmon jumping at an angle of 31.4 ◦ leave the water to continue upstream?

Question

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.73 m away from a waterfall 0.504 m in height, at what minimum speed must a salmon jumping at an angle of 31.4 ◦ 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

Interesting problem. Here is the approach I took:

First, draw a picture of the initial velocity:

Looking at the x-direction: x = v_xt but v_x = vcos(31.4°) so x = vcos(31.4°)t or 2.73 = vcos(31.4°)t which means that t = 2.73/[vcos(31.4°)]

Looking at the y direction:

If we look at the motion it would (could) do something like this:

Point 2 (P₂) is farther than P₁ so let's assume that is where the fish hits the water.

In the y direction, Δy = v_yit + 1/2at² but v_yi = vsin(31.4°) so 0.504 = vsin(31.4°)t +1/2(-9.81)t². We can now substitute in (from the x-direction work) t = 2.73/[vcos(31.4°)] so we get:

0.504 = vsin(31.4°)(2.73/[vcos(31.4°)]) +1/2(-9.81)(2.73/[vcos(31.4°)])² or

0.504 = 2.73/tan(31.4°) - 50.1771/v²

so v² = 43.1668 or v = 6.57 m/s

To check out our answer, we can do a couple things. First, plug in v = 6.57 into 0.504 = vsin(31.4°)t +1/2(-9.81)t² to see what the 2 times are for the fish to reach 0.504 meters. It results in the quadratic equation -4.905t² + 3.423t - 0.504 and solving, we get t = 0.211s and 0.487s which is consistent with our answer.

Next, we can look at the the distance traveled in both the x and y directions in t = 0.487s and it seems consistent with our problem so I think this looks good.