Respuesta :
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
[tex]mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2[/tex]
[tex]mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2[/tex]
[tex]mgh_i = \dfrac{1}{2}mv_f^2[/tex]
[tex]v_f = \sqrt{2gh_i}[/tex]
[tex]= \sqrt{2gL(1- cos\theta)}[/tex]
[tex]= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}[/tex]
= 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
[tex]mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2[/tex]
[tex]mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2[/tex]
[tex]mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2[/tex]
[tex]gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2[/tex]
[tex]v_f = \sqrt{v_1^2+2gh_i}[/tex]
[tex]v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}[/tex]
v_f= 11.29 m/s
