While watching the clouds pass by, you notice a European swallow flying horizontally at a height h = 21.4 m above you. When the swallow is directly overhead, it drops an m = 11.1 kg coconut. From your ornithological studies, you know that the air-speed of this particular species of swallow while carrying such a load is v0 = 6.61 m/s. In this problem you may neglect air friction. Assume the origin of the coordinate system there is at your head. Calculate the magnitude of angular momentum for the coconut as observed by you after one second after it is released directly overhead,

Respuesta :

Answer:

magnitude of angular momentum of the coconut after t = 1 s is

[tex]L = 1930.4 kg m^2/s[/tex]

Explanation:

Since it is dropped at the instant when it is at overhead position

so we will have

[tex]v_x = 6.61 m/s[/tex]

now vertical speed of the coconut after t = 1 s

[tex]v_y = 0 - gt[/tex]

[tex]v_y = -9.8 m/s[/tex]

now its displacement in x direction is given as

[tex]x = 6.61 (1) = 6.61 m[/tex]

height of the coconut after t = 1 s

[tex]y = 21.4 - \frac{1}{2}gt^2[/tex]

[tex]y = 21.4 - \frac{1}{2}(9.81)(1^2)[/tex]

[tex]y = 16.5 m[/tex]

now angular momentum is given as

[tex]L = \vec r \times m\vec v[/tex]

now we have

[tex]L = (6.61 \hat i + 16.5 \hat j)\times (11.1)(6.61 \hat i - 9.81\hat j)[/tex]

[tex]L = -719.8 \hat k - 1210.6\hat k[/tex]

so magnitude of angular momentum of the coconut after t = 1 s is

[tex]L = 1930.4 kg m^2/s[/tex]

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