Answer:
The velocity at the bottom of the swing is 9.71 m/s
Solution:
As per the question:
Length of the vine, l = 30.0 m
Incident angle, [tex]\theta = 32.9^{\circ}[/tex]
Speed, v = 3.30 m/s
Now,
From Fig. 1
AD = l
AB = [tex]lcos\theta [/tex]
AC = [tex]l - lcos\theta = l(1 - cos\theta)[/tex]
To calculate the speed at the bottom of the swing:
Initially at rest, i.e., initial velocity, v = 0 m/s
Potential Energy, [tex]PE = mgl - mglcos\theta = mgl(1 - cos\theta)[/tex]
At the bottom of the swing:
Kinetic Energy, [tex]KE = \frac{1}{2}mv_{b}^{2}[/tex]
Now, by using the law of conservation of energy:
PE = KE
[tex]mgl(1 - cos\theta) = \frac{1}{2}mv'^{2}[/tex]
[tex]v_{b} = \sqrt{2gl(1 - cos\theta)}[/tex]
Substituting the appropriate value in the above eqn:
[tex]v_{b} = \sqrt{2\times 9.8\times 30.0\times (1 - cos32.9)} = 9.71\ m/s[/tex]
