Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00m from a waterfall that is 0.55m tall and ju

Question

Luba_88

14 days ago

8

Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00m from a waterfall that is 0.55m tall and ju

mps at an angle of 32.0 degrees. What must be the salmon's minimum speed to reach the waterfall?

Mathematics

2 answers:

babunello [3.6K] · Answer 1 · 2026-02-19T00:13:33+03:00

Answer: 6.2 m/s

Explanation:

1) This situation involves projectile motion, which follows a parabolic path.

2) Velocity components:

i) Initial velocity is denoted as V₀.

ii) Horizontal velocity component:

V₀x = V₀ cos α

Since horizontal velocity remains constant, Vx equals V₀x.

iii) Vertical velocity component:

V₀y = V₀ sin α

Its change over time is affected by acceleration due to gravity, g ≈ 9.8 m/s².

Vy = V₀y - gt = V₀ sin α - gt

3) Formulas for displacement:

i) Horizontal displacement:

x = V₀ cosα × t

ii) Vertical displacement:

y = V₀ sin α × t - (1/2) g t²

or y ≈ V₀ sin α × t - 4.9 t²

4) Working out the solution:

i) We know x = V₀ cos(32°) t = 2.00 m, because the salmon begins 2.00 m away from the waterfall.

Thus, V₀ = 2 / [cos(32°) × t]

ii) For vertical displacement: y = V₀ sin(32°) t - 4.9 t²

iii) Substituting V₀, y = sin(32°) × 2 / [cos(32°) × t] × t - 4.9 t² = 2 tan(32°) - 4.9 t²

iv) Setting y = 0.55 m (the height of the waterfall) to solve for t:

0.55 = 2 tan(32°) - 4.9 t²

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

t = √0.143 ≈ 0.38 s

v) Finally, compute V₀:

V₀ = 2 / [cos(32°) × 0.38] = 6.2 m/s

Leona [4.1K] · Answer 2 · 2026-02-19T00:10:34+03:00

The salmon must have a minimum velocity of 6.24 m/s to successfully reach the waterfall.

Additional Details

Acceleration is defined as the change rate of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m/s²)

v = final velocity (m/s)

u = initial velocity (m/s)

t = time elapsed (s)

d = distance covered (m)

Let's solve this problem!

This relates to Projectile Motion.

Given data:

Horizontal distance, x = 2.00 m

Vertical distance, y = 0.55 m

Launch angle, θ = 32.0°

Unknown:

Initial speed, u = ?

Solution steps:

$x = u \times \cos \theta \times t$

$t = \boxed{\frac{x}{u \cos \theta}}$ (Equation 1)

$y = (u \times \sin \theta \times t) - (\frac{1}{2}\times g \times t^2)$

$y = (u \times \sin \theta \times \frac{x}{u \cos \theta}) - (\frac{1}{2}\times g \times (\frac{x}{u \cos \theta})^2)$ (Using Equation 1)

$y = (x \tan \theta) - (\frac{gx^2}{2u^2 \cos ^2 \theta})$

$0.55 = (2.00 \tan 32^o) - (\frac{9.8 (2.00)^2}{2u^2 \cos ^2 32^o})$

$0.55 = 1.25 - \frac{27.3}{u^2}$

$\frac{27.3}{u^2} = 1.25 - 0.55$

$u^2 = \frac{27.3}{1.25 - 0.55}$

$u = \sqrt {\frac{27.3}{1.25 - 0.55}}$

$u = \sqrt {39}$

$u \approx \boxed {6.24 ~ \text{m/s}}$

Answer specifics

Grade level: High School

Subject area: Physics

Topic: Kinematics

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

Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00m from a waterfall that is 0.55m tall and ju

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