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 0.70 kg squid is able to eject 0.26 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: V=7.43m/s

d = 2.82m

Explanation:

a) For the first part, the initial velocity immediately after ejection, by using momentum conservation

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

there are no external forces acting on the system at the moment of ejection, so we can find the speed of the squid by noting

momentum before ejection = momentum after ejection

0 = M1U + M2V

0=-0.26 kg x 20 m/s + 0.7kg x 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=7.43m/s

B. we use the equation vf^²=v0^²+2ad

where vf=final velocity = 0 since velocity is zero at motion's apex

v0=initial velocity = 7.43m/s

a = acceleration = - 9.8m/s/s

d=height (to be found)

Therefore,

0=7.43^²+2 (-9.8) d

Mathematically, it becomes

d=7.43^²/2 (9.8) = 2.82m

d = 2.82m