Answer:
The momentum of the bale the moment it strikes the ground is 1076.68 kg-m/s.
Explanation:
It is given that,
Velocity of an animal-rescue plane, [tex]v_x=55\ m/s[/tex]
It drops a bale of hay from an altitude of 69 m, h = 69 m
The vertical velocity of plane is given by :
[tex]v_y=\sqrt{2gh}[/tex]
[tex]v_y=\sqrt{2\times 9.81\times 69} =36.79\ m/s[/tex]
Weight of the bale of hay, W = 192 N
If m is the mass, then weight is given by :
[tex]m=\dfrac{W}{g}[/tex]
[tex]m=\dfrac{192}{9.81}=19.57\ kg[/tex]
The resultant momentum of the bale the moment it strikes the ground is given by :
[tex]p=m(v_x-v_y)[/tex]
[tex]p=19.57\times 55i-19.57\times 36.79j[/tex]
[tex]p=(1076.35i-719.98j)\ kg-m/s[/tex]
Magnitude of momentum,
[tex]p=\sqrt{1076.35^{2}+719.98}[/tex]
p = 1076.68 kg-m/s
So, the momentum of the bale the moment it strikes the ground is 1076.68 kg-m/s. Hence, this is the required solution.