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Tarzan swings on a 36.0 m long vine initially inclined at an angle of 31.0◦ with the vertical. The acceleration of gravity if 9.81 m/s 2 . What is his speed at the bottom of the swing if he a) starts from rest? Answer in units of m/s.

Respuesta :

Answer:

  v = 10 m/s

Explanation:

Given,

Length of the vine, l = 36 m

Angle of inclination = 31.0◦ with the vertical

acceleration due to gravity = 9.81 m/s²

Using Conservation of energy

      KE = PE

 [tex]\dfrac{1}{2} mv^2 = m g h[/tex]

[tex] v = \sqrt{2gh}[/tex]

[tex] v = \sqrt{2 g l(1-\cos \theta)[/tex]

[tex] v = \sqrt{2\times 9.81\times 36 \times (1-\cos 31^0)[/tex]

  v = 10 m/s

Hence, the speed of the swing is equal to 10 m/s

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