Question

Salmon often jump waterfalls to reach their breeding grounds starting downstream, 2.9 meters away from a waterfall .436 meters in height, at what minimum speed must a Salmon jumping at an angle of 44.7 degrees leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s^2

Answer 1

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 $u_{x}$ = u cosФ

2. Vertical component $u_{y}$ = u sinФ

where u is the initial velocity and Ф is the angle between the horizontal

and the initial velocity u

→ Ф = 44.7°

→ $u_{x}$ = u cos(44.7)

→ $u_{y}$ = 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 = $u_{x}$ t

4. The vertical distance y = $u_{y}$ t + $\frac{1}{2}$ 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 = $\frac{2.9}{ucos(44.7)}$ ⇒ (1)

→ y = u sin(44.7) t + $\frac{1}{2}$ 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)($\frac{2.9}{ucos(44.7)}$) - 4.905 ($\frac{2.9}{ucos(44.7)}$)²

→ 0.436 = 2.9 ($\frac{sin(44.7)}{cos(44.7)}$ - $\frac{41.25105}{u^{2}[cos(44.7)]^{2}}$

→ 0.436 = 2.8698 - $\frac{81.4671}{u^{2} }$

Subtract 2.8698 from both sides

→ -2.4338 = - $\frac{81.4671}{u^{2} }$

Multiply both sides by -1

→ 2.4338 =  $\frac{81.4671}{u^{2} }$

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:

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