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 cosФ
2. Vertical component 
= 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 cos(44.7)
→ = 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 = t
4. The vertical distance y = t + 
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 = 
⇒ (1)
→ y = u sin(44.7) t + 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)() - 4.905 ()²
→ 0.436 = 2.9 (
- ![\frac{41.25105}{u^{2}[cos(44.7)]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B41.25105%7D%7Bu%5E%7B2%7D%5Bcos%2844.7%29%5D%5E%7B2%7D%7D)
→ 0.436 = 2.8698 - 
Subtract 2.8698 from both sides
→ -2.4338 = -
Multiply both sides by -1
→ 2.4338 =
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
<em>The minimum speed must a Salmon jumping with to leave the water</em>
<em>to continue upstream is 5.79 m/s </em>
Learn more:
You can learn more about the equation of trajectory of the projectile in brainly.com/question/2814900
brainly.com/question/5531630
#LearnwithBrainly
<em> = 700 years</em>