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 ([tex]g=-9.8 m/s^2[/tex], acceleration of gravity)
The horizontal velocity of the salmon is given by:
[tex]v_x = u cos \theta[/tex]
where
u = ? is the initial speed
[tex]\theta=32^{\circ}[/tex] is the angle of projection
Then the horizontal distance covered by the salmon after a time t is given by
[tex]d=v_x t =(u cos \theta) t[/tex]
Or equivalently, the time taken to cover a distance d is
[tex]t=\frac{d}{u cos \theta}[/tex] (1)
Along the vertical direction, the equation of motion is
[tex]h = (u sin \theta) t + \frac{1}{2}gt^2[/tex] (2)
where
[tex]u sin \theta[/tex] is the initial vertical velocity
If we substitute (1) into (2), we get:
[tex]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}[/tex]
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:
[tex]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[/tex]
