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]