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The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at [tex]v_{i,2} = 17.9 \, \text{m/s}[/tex] when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle [tex]\theta = 61.9^\circ[/tex] below the horizontal and a speed of [tex]v_{i,1} = 35.9 \, \text{m/s}[/tex].

What is the speed of the eagle immediately after it catches its prey?

Answer :

Answer:

V=27.24 m/s

Explanation:

We need to apply the linear momentum conservation theorem:

[tex]m_1*v_{o1}+m_2*v_{o2}=m_t*v_{f}\\[/tex]

The velocity of the eagle its defined by its two components:

[tex]v_x=V*cos(\theta)\\v_y=V*sin(\theta)\\v_x=35.9*cos(61.9^o)=16.9m/s\\v_y=35.9*sin(61.9^o)=31.7m/s[/tex]

[tex]2*(16.9m/s(i)+31.66m/s(j))+17.9m/s(i)=3*v_f\\v_f=17.23m/s(i)+21.10m/s(j)[/tex]

because speed is a scalar value:

[tex]V=\sqrt{(21.10m/s)^2+(17.23)^2}\\V=27.24m/s[/tex]

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