Question

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom of the swing if he starts from rest? m/s (b) What is his speed at the bottom of the swing if he starts with an initial speed of 3.00 m/s? m/s

Answer 1

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

       $mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

          $mgh_i = \dfrac{1}{2}mv_f^2$

             $v_f = \sqrt{2gh_i}$

                   $= \sqrt{2gL(1- cos\theta)}$

                   $= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

       $mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

          $mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

          $gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

             $v_f = \sqrt{v_1^2+2gh_i}$

             $v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

                       v_f= 11.29 m/s