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Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00m from a waterfall that is 0.55m tall and jumps at an angle of 32.0 degrees. What must be the salmon's minimum speed to reach the waterfall?

Respuesta :

Edufirst

Answer: 6.2 m/s

Explanation:

1) This is a projectile motion (parabolic)

2) Velocity:

i) initial velocity = V₀

ii) Horizontal component:

V₀x = V₀ cos α

The horizontal velocity is constant, so Vx = V₀x

ii) Vertical component:

V₀y = V₀ sin α

The vertical component is linear with acceleration = g ≈ 9.8 m/s²

Vy = V₀y - gt = V₀ sin α - gt

3) Displacement equations

i) Horizontal displacment:

x = V₀ cosα t

ii) Vertical displacement:

y = V₀ sin α t - g t² / 2

y ≈ V₀ sin α t - 4.9 t²

4) Solution

i) x = V₀ cosα t = V₀ cos(32°) t = 2.00 m ← salmon starts 2.00m from a waterfall

⇒ V₀ = 2 / [cos(32°) t ]

ii) y = V₀ sin α t - 4.9 t² = V₀ sin(32°) t - 4.9 t²

iii) Replace V₀ with 2 / [cos(32°) t ]

y = sin(32°) × 2 / [cos(32°) t ] × t - 4.9t² = 2 tan(32°) - 4.9t²

iv) Use jump's height (y = 0.55m) and solve

⇒ 0.55 = 2 tan(32°) - 4.9t²

t² = [2 tan(32°) - 0.55 ] / 4.9 = 0.143 s²

⇒ t = √ (0.143s²) = 0.38 s

v) Use V₀ = 2 / [cos(32°) t ] to find V₀

V₀ = 2 / [cos(32°) (0.38) ] = 6.2 m/s

johanrusli

The salmon's minimum speed must be 6.24 m/s to reach the waterfall

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

This problem is about Projectile Motion

Given:

Horizontal distance = x = 2.00 m

Vertical distance = y = 0.55 m

Angle of projection = θ = 32.0°

Unknown:

Initial speed = u = ?

Solution:

[tex]x = u \times \cos \theta \times t[/tex]

[tex]t = \boxed{\frac{x}{u \cos \theta}}[/tex] → Equation 1

[tex]y = (u \times \sin \theta \times t) - (\frac{1}{2}\times g \times t^2)[/tex]

[tex]y = (u \times \sin \theta \times \frac{x}{u \cos \theta}) - (\frac{1}{2}\times g \times (\frac{x}{u \cos \theta})^2)[/tex] ← Equation 1

[tex]y = (x \tan \theta) - (\frac{gx^2}{2u^2 \cos ^2 \theta})[/tex]

[tex]0.55 = (2.00 \tan 32^o) - (\frac{9.8 (2.00)^2}{2u^2 \cos ^2 32^o})[/tex]

[tex]0.55 = 1.25 - \frac{27.3}{u^2}[/tex]

[tex]\frac{27.3}{u^2} = 1.25 - 0.55[/tex]

[tex]u^2 = \frac{27.3}{1.25 - 0.55}[/tex]

[tex]u = \sqrt {\frac{27.3}{1.25 - 0.55}}[/tex]

[tex]u = \sqrt {39}[/tex]

[tex]u \approx \boxed {6.24 ~ \text{m/s}}[/tex]

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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