Question

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 6.65 m/s at 48.1° 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 3.79 m/s before jumping out again. (a) determine the average velocity of the fish for the entire process of jumping and swimming underwater. incorrect: your answer is incorrect. your response differs from the correct answer by more than 10%. double check your calculations. m/s (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 3.79 m/s?

Answer 1

When it jumps, the fish travels a horizontal distance of

`d_1=(v^2/g)*sin(2\theta) =(6.65^2/9.81)*\sin(2*48.1) =4.48 m`
The horizontal speed when the fish jumps is

`v_1=v\cos \theta =6.65*\cos(48.1) =4.44 (m/s)`
The average speed is give by the equation

`v=(v_1*d_1+v_2*d_2)/(d_1+d_2)`

`v=(4.44*4.48 +3.79*4.48)/(4.48+4.48)=4.115 (m/s)`
Time to travel this way (jumping and swimming) is

`t=t_1+t_2=4.48/4.44+4.48/3.79 =2.19 s`
If it were to swim the time was

`tt=2*t_2=2*4.48/3.79 =2.364 s`
Percentage change is

`\epsilon =(tt-t)/t =(2.364-2.19)/2.364 =0.073=7.3%`