Question

A flying squid (family Ommastrephidae) is able to "jump" off the surface of the sea by taking water into its body cavity and then ejecting the water vertically downward. A 0.74-kg squid is able to eject 0.33 kg of water with a speed of 20 m/s.

(a) What will be the speed of the squid immediately after ejecting the water?

(b) How high in the air will the squid rise?

Answer 1

a)
By using momentum conservation

before ejection, the momentum of the squid/water system is zero

Momentum before ejection = Momentum after ejection
0= - 0.33 kg x 20 m/s + 0.74 kg * V

where the speed of the water is taken as the negative sign, and V is the speed of the squid right after ejection, solving for V we get
V = 8.92 m/s
Speed of the squid immediately after ejecting the water, V = 8.92 m/s

b)
vf^2=v0^2 + 2ad

where vf=final velocity =0
v0=initial velocity = 8.92 m/s
a=accel = - 9.8m/s^2
d=height

substituing Values =

d = 8.92^2/(2*9.8)
d = 4.1 m
High in the air will the squid rise, d = 4.1 m