Salmon often jump waterfalls to reach their breeding grounds.

Starting downstream, 1.88 m away from a waterfall 0.262 m in height, at what minimum speed must a salmon jumping at an angle of 43.5° leave the water to continue upstream?

The acceleration due to gravity is 9.81 m/s^2.

Answer in units of m/s.

4.65 m/s

Explanation:

Given:

θ = 43.5°

Δx = 1.88 m

Δy = 0.262 m

aₓ = 0 m/s²

aᵧ = - 9.81 m/s²

Find: v₀

Δx = v₀ₓ t + ½ aₓ t²

(1.88 m) = (v₀ cos 43.5°) t + ½ (0 m/s²) t²

t = 1.88 m / (v₀ cos 43.5°)

t = 2.59 m / v₀

Δy = v₀ᵧ t + ½ aᵧ t²

(0.262 m) = (v₀ sin 43.5°) t + ½ (-9.81 m/s²) t²

Substitute:

0.262 m = (v₀ sin 43.5°) (2.59 m / v₀) + ½ (-9.81 m/s²) (2.59 m / v₀) ²

0.262 m = 2.59 m sin 43.5° - 32.9 m³/s² / v₀²

v₀ = 4.65 m/s