. Chinook salmon are able to move through water especially
fast by jumping out of the water periodically. This behavior is called porpoising. Suppose a salmon swimming in
still water jumps out of the water with velocity 6.26 m/s at
45.0° 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?
Answers
Answer:
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Explanation:
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Answer:
a) The vertical component of the salmon’s velocity as it leaves the water is
v
yi
=+v
i
sinθ=+(6.26m/s)sin45.0
0
≃+4.43m/s
When the salmon returns to water level at the end of the leap, the vertical component of velocity will be
v
yf
=−v
yi
≃−4.43m/s
If the salmon jumps out of the water at t=0, the time interval required for it to return to the water is
Δ
1
=
a
y
v
yf
−v
yi
=
−9.80m/s
2
−4.43m/s−4.43m/s
=0.903s
The horizontal distance traveled during the leap is
L=v
xi
Δt
i
=(v
i
cosθ)θt
i
=(6.26m/s)cos45.0
0
(0.903s)=4.00m
To travel this same distance underwater, at speed v=3.58m/s, requires a time interval of
Δt
2
=
v
L
=
3.58m/s
4.00m
≃1.12s
The average horizontal speed for the full porpoising maneuver is then
v
avg
=
Δt
Δx
=
Δt
1
+Δt
2
2L
=
0.903s+1.12s
2(4.00m)
=3.96m/s
(b) From (a), the total time interval for the porpoising maneuver is
Δt=0.903s+1.12s=2.02s
Without porpoising, the time interval to travel distance 2L is
Δt
2
=
v
2L
=
3.58m/s
8.00m
≃2.23s
The percentage difference is
Δt
2
Δt
1
−Δt
2
∗100% =−9.6%
Porpoising reduces the time interval by 9.6%.