Question

Chinook salmon can cover more distance in less time by periodically making jumps out of the water. Suppose a salmon swimming in still water jumps out of the water with velocity 5.63 m/s at 46.6° above the horizontal, re-enters the water a distance L upstream, and then swims the same distance L underwater in a straight, horizontal line with velocity 2.32 m/s before jumping out again.

a) What is the fish's average horizontal velocity (in m/s) between jumps? (Round your answer to at least 2 decimal places.)

b) What If? Some salmon are able to jump a distance L out of the water while only swimming a distance

L

4

between jumps. By what percentage are these salmon faster than those requiring an underwater swim of distance L? (Assume the salmon jumps out of the water with velocity 5.63 m/s at 46.6° above the horizontal, re-enters the water a distance L upstream, and then swims a distance

L

4

underwater in a straight, horizontal line with velocity 2.32 m/s before jumping out again.)

Answer 1

a] Total time taken = jumping time + swimming time

= L/(5.63 cos 46.6 degree) + L/2.32

average horizontal velocity = Total distance/total time

= 2L /[L/(5.63 cos 46.6 degree) + L/2.32]

= 2/[1/(5.63 cos 46.6 degree) + 1/2.32]

= 2/0.6895

= 2.9 m/s answer

b] average horizontal velocity = Total distance/total time

= (L+L/4) /[L/(5.63 cos 46.6 degree) + (L/4)/2.32]

= 1.25/[1/(5.63 cos 46.6 degree) + 1/9.28]

= 3.413 m/s

percentage faster = [3.413-2.9]*100/2.9

= 17.7 % answer