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
