The minimum speed must a Salmon jumping with to leave the water
to continue upstream is 5.79 m/s
Explanation:
At first let us find the two component of the jumping velocity of the fish
1. Horizontal component [tex]u_{x}[/tex] = u cosФ
2. Vertical component [tex]u_{y}[/tex] = u sinФ
where u is the initial velocity and Ф is the angle between the horizontal
and the initial velocity u
→ Ф = 44.7°
→ [tex]u_{x}[/tex] = u cos(44.7)
→ [tex]u_{y}[/tex] = u sin(44.7)
The horizontal distance x is 2.9 meters away from a waterfall
The vertical distance y is 0.436 meters
3. The horizontal distance x = [tex]u_{x}[/tex] t
4. The vertical distance y = [tex]u_{y}[/tex] t + [tex]\frac{1}{2}[/tex] gt²
where g is the acceleration of gravity
→ x = u cos(44.7) t
→ x = 2.9 meters
→ 2.9 = u cos(44.7) t
Divided both sides by u cos(44.7)
→ t = [tex]\frac{2.9}{ucos(44.7)}[/tex] ⇒ (1)
→ y = u sin(44.7) t + [tex]\frac{1}{2}[/tex] gt²
→ y = 0.436 meters , g = -9.81 m/s²
→ 0.436 = u sin(44.7) t - 4.905 t² ⇒ (t)
Substitute (1) in (2) to make the equation of u only
→ 0.436 = u sin(44.7)([tex]\frac{2.9}{ucos(44.7)}[/tex]) - 4.905 ([tex]\frac{2.9}{ucos(44.7)}[/tex])²
→ 0.436 = 2.9 ([tex]\frac{sin(44.7)}{cos(44.7)}[/tex] - [tex]\frac{41.25105}{u^{2}[cos(44.7)]^{2}}[/tex]
→ 0.436 = 2.8698 - [tex]\frac{81.4671}{u^{2} }[/tex]
Subtract 2.8698 from both sides
→ -2.4338 = - [tex]\frac{81.4671}{u^{2} }[/tex]
Multiply both sides by -1
→ 2.4338 = [tex]\frac{81.4671}{u^{2} }[/tex]
By using cross multiplication
∴ 2.4338 u² = 81.4671
Divide both sides by 2.4338
→ u² = 33.4732
Take √ for both sides
→ u = 5.79 m/s
The minimum speed must a Salmon jumping with to leave the water
to continue upstream is 5.79 m/s
Learn more:
You can learn more about the equation of trajectory of the projectile in brainly.com/question/2814900
brainly.com/question/5531630
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