Question

Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 3.17 m away from a
waterfall 0.432 m in height, at what minimum
speed must a salmon jumping at an angle of
30.8º leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s?.
Answer in units of m/s.

Answer 1

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

• height of the waterfall, h = 0.432 m
• distance of the Salmon from the waterfall, s = 3.17 m
• angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

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