Question

Salmon often jump waterfalls to reach their

Answer 1

Answer:

6.35 m/s

Explanation:

The motion of the salmon is equivalent to that of a projectile, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration ($g=-9.8 m/s^2$, acceleration of gravity)

The horizontal velocity of the salmon is given by:

$v_x = u cos \theta$

where

u = ? is the initial speed

$\theta=32^{\circ}$ is the angle of projection

Then the horizontal distance covered by the salmon after a time t is given by

$d=v_x t =(u cos \theta) t$

Or equivalently, the time taken to cover a distance d is

$t=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$h = (u sin \theta) t + \frac{1}{2}gt^2$ (2)

where

$u sin \theta$ is the initial vertical velocity

If we substitute (1) into (2), we get:

$h = (u sin \theta) \frac{d}{cos \theta} + \frac{1}{2}g(\frac{d}{ ucos \theta})^2=d tan \theta + \frac{gd^2}{2u^2 cos^2 \theta}$

We now that in order to reach the breeding grounds, the salmon must travel a distance of

d = 2.02 m

reaching a height of

h = 0.574 m

Substituting these data into the equation and solving for u, we find the initial speed that the salmon must have:

$u =\sqrt{ \frac{gd^2}{2(h-d tan \theta) cos^2 \theta}}=\sqrt{\frac{(-9.8)(2.02)^2}{2(0.574-(2.02)(tan 32))(cos^2(32))}}=6.35 m/s$