Question

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?

Answer 1

V=27.24 m/s

Step-by-step 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=√((21.10m/s)^2+(17.23)^2)\\V=27.24m/s`