A salmon swimming in still water jumps out of the water with velocity 6.26 m/s a

Question

A salmon swimming in still water jumps out of the water with velocity 6.26 m/s at 45° above the horizontal, sails through the air a distance L before returning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.58 m/s before jumping out again. (a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater. (b) Consider the time interval required to travel the entire distance of 2L. By what percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater at 3.58 m/s?

Explanation / Answer

Salmon's velocity has two components, v[y] and v[x], we can get them by decomposing v.
So
cos(45º) = v[x]/v
v[x] = v*2/2 = 3.13*2 m/s

sin(45º) = v[y]/v
v[y] = v*2/2 = 3.13*2 m/s

In y:
Using v = v0 + g*t
When it reaches its max height v = 0
g*t = v0
t = v0/g
Assuming g = 10 m/s^2
t = 0.313*2 s

The time it takes to reaches the top is the same to reaches the water again. So the total time is
0.626*2 s.

In x:
v[x] = L/t, where t = 0.626*2 s
3.13*2 = L/(0.626*2)
L = 3.92 m

Now it's swimming with velocity 3.58 m/s and it's going to swims the same distance L.
v = L/t
3.58 = 3.92/t
t = 1.09 s

(a)
Its average velocity is s = 2*L = 7.84 m and the time is t + t = 0.626*2 + 1.09 = 1.98 s
v = s/t = 7.84/1.98 = 3.96 m/s

(b)
Only swimming:
v = 2*L/t
t = 7.84/3.58 = 2.19 s

So, 1.98/2.19 = 0.904
It reduces in 10% the time.

Ok?

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