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Lapatulllka [165]

3 years ago

7

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ????????,2=17.9vi,2=17.9

m/s 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 theta=61.9θ=61.9 ° below the horizontal and a speed of ????????,1=35.9vi,1=35.9 m/s. What is the speed of the eagle immediately after it catches its prey?

1 answer:

AleksAgata [21]3 years ago

6 0

Answer:

V=27.24 m/s

Explanation:

We need to apply the linear momentum conservation theorem:

$m_1*v_{o1}+m_2*v_{o2}=m_t*v_{f}\\$

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

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

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

because speed is a scalar value:

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

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3 years ago

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2 years ago

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