Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 1.87 m away from a
waterfall 0.37 m in height, at what minimum
speed must a salmon jumping at an angle of
32.7◦ leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s
Answer in units of m/s.

Let [tex]v\rm \; m\cdot s^{-1}[/tex] be the minimum speed at which that salmon leaves the water with an elevation angle of [tex]32.7^{\circ}[/tex]. [tex]v > 0[/tex].

Initial horizontal velocity: [tex]v\; \cos{32.7^{\circ}}[/tex];
Initial vertical velocity: [tex]v \; \sin{32.7^{\circ}}[/tex].

Why is the initial horizontal velocity important? If there's no air resistance, the horizontal velocity of the fish will stay the same during its flight.

By the time the fish reaches the upper rim of the waterfall, it would have traveled [tex]\rm 1.87\; m[/tex] horizontally. How much time after take-off will that occur?

[tex]\displaystyle t = \frac{s_\text{horizontal}}{v_\text{horizontal}} = \frac{1.87}{v\; \cos{32.7^{\circ}}}[/tex].

For the fish to continue upstream, its height should be at least [tex]\rm 0.37\; m[/tex] by the time it reaches the upper rim of the waterfall. In other words, at [tex]\displaystyle t = \frac{1.87}{v\; \cos{32.7^{\circ}}}[/tex], [tex]h \ge 0.37[/tex].

If there's no air resistance, on the vertical direction the fish would accelerates downwards at a constant [tex]g =\rm -9.81\; m\cdot s^{-1}[/tex] during the entire flight. Its height could be expressed as

[tex]\displaystyle h = \frac{1}{2}g \cdot t^{2} + v_{\text{vertical,} \atop \text{initial}}\cdot t[/tex].

That's the same as:

[tex]\begin{aligned} h &= \frac{1}{2}g \cdot t^{2} + v_{\text{vertical,} \atop \text{initial}}\cdot t\\ &= \frac{1}{2}(-9.81) \cdot \left(\frac{1.87}{v\; \cos{32.7^{\circ}}}\right)^{2} + (v\; \cos{32.7^{\circ}})\cdot \frac{1.87}{v\; \cos{32.7^{\circ}}} \end{aligned}[/tex].

Let [tex]h = \rm 0.37\; m[/tex], Solve this equation for [tex]v[/tex]

[tex]\displaystyle \frac{1}{2}(-9.81) \cdot \left(\frac{1.87}{v\; \cos{32.7^{\circ}}}\right)^{2} + (v\; \cos{32.7^{\circ}})\cdot \frac{1.87}{v\; \cos{32.7^{\circ}}} = 0.37[/tex].

[tex]\displaystyle \frac{1}{2}(-9.81) \cdot \left(\frac{1.87}{v\; \cos{32.7^{\circ}}}\right)^{2} + 1.87= 0.37[/tex]

Given that [tex]v > 0[/tex], [tex]v = 5.40[/tex].

In other words, the fish must leave the water at a speed of at least [tex]\rm 5.40\; m\cdot s^{-1}[/tex] to continue upstream.