Question

A Chinook Salmon has a maximum underwater speed of 3.58 m/s, but itcan jump out of water with a speed of 6.08 m/s. To move upstream past a waterfall, thesalmon doesnot need to jump to the top of the fall, but only to apoint in the fall where the water speed is less than 3.58 m/s; itcan then swim up the fall for the remainingdistance. Because thesalmon must make forward progress in the water, let's assume thatit can swim to the top if the water speed is 3.00 m/s. Assume waterhas a speed of1.80 m/s as it passesover a ledge.(a) How far below the ledge will the water bemoving with a speed of 3.00 m/s? (Note that water undergoesprojectile motion once it leaves the ledge.)
m

(b)If the salmon is able to jump vertically upward from the base ofthe fall, what is the maximum height of waterfall that the salmoncan clear?
m

Answer 1

a)
final velocity of water is V_f =3 m/s
initial velocity of water is V_i= 1.5 m/s
acceleration due to gravity isg = 9.8m/s


we have from the equation of motion we have
V_f²= V_i² +2aΔy

orΔy=(V_f²-V_i²) / 2gcalculatefor Δy


b)here

V_i = velocity atthe instant right before the jump = 0 m/s
V_f= Velocity of fish jumpforce-resistance of water= 6.08-1.5= 4.58 m/s
a = -9.8 m/s²(since the y-axis varies in this one it is important to keepnegitive sign)again we useΔy= (V_f²-V_i²) / 2gcalculate for Δy