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

(a) In the air: initial vertical speed=6.26*sin45.0 = 4.426 m/s vertical acceleration, g = 9.8 m/s^2 vertical displacement = 0 time spend in the air = t From equations of motion, h = u*t+(1/2)*g*t2 0 = 4.426 - (1/2)*9.8*t2 t = 0.95 s horizontal velocity=6.26*cos45 = 4.426 m/s horizontal acceleration=0 L = 4.426*0.95 L = 4.2 m under water time taken to travel L,t = 4.2/3.58=1.17 s total time taken to travel 2*L=1.17 + 0.95= 2.12 s therefore the velocity = distance/time = (2*L)/1.97 = 2*4.2/2.12 = 3.95 m/s ( b) the required persentage is = [2.12 - 1.17]* 100 PLEASE RATE.

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