Question

Tarzan swings on a 30.0-m-long vine initially inclined at an angle of 37 degrees measured from vertical. What is his speed at the bottom of the swings if he starts from rest?

Answer 1

Answer:

10.8 m/s

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the gravitational potential energy at the beginning of the motion will be equal to the kinetic energy at the bottom of the swing:

$U=K\\mgh=\frac{1}{2}mv^2$

where

m is Tarzan's mass

g = 9.8 m/s^2 is the gravitational acceleration

h is the initial heigth of Tarzan

v is the speed of Tarzan at the bottom

Re-arranging the equation for v, we have

$v=\sqrt{2gh}$

Therefore, we need to find h, the initial height of Tarzan. We know that the length of the swing is L=30.0 m and it is initially inclined at $\theta=37^{circ}$ with respect to the vertical, so the initial heigth of Tarzan is given by

$h=L-Lcos\theta =L(1-cos \theta)=(30.0 m)(1-cos 37^{\circ})=6.0 m$

And so, Tarzan's speed at the bottom of the swing is

$v=\sqrt{2(9.8 m/s^2)(6.0 m)}=10.8 m/s$