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Sagot :
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;
[tex]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[/tex]
The minimum velocity of the Salmon jumping at the given angle is calculated as;
[tex]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[/tex]
Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.
Learn more here: https://brainly.com/question/20064545
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