Question

Tarzan swings on a 31.0-m-long vine initially inclined at an angle of 42.0° with the vertical. What is his speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s? (Pick the answer closest to the true value.)A. 13.9 m/sB. 12.5 m/sC. 8.4 m/sD. 6.9 m/sE. 11.0 m/s

Answer 1

Answer:

Speed of Tarzan at the bottom of the swing is 12.5 m/s.

Explanation:

Given that,

Length of the vine, L = 31 m

The swing is inclined at an angle of 42 degrees with the vertical. We need to find the speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s. using the conservation of mechanical energy as :

$mgh=\dfrac{1}{2}mv^2$

$gh=\dfrac{1}{2}v^2$

h is the height of the height.

$h=L-L\ cos\theta$

$v=\sqrt{2gh}$

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

$v=\sqrt{2\times 9.8\times 31(1-\ cos(42))}$

v = 12.492 m/s

or

v = 12.5 m/s

So, his speed at the bottom of the swing if he pushes off with a speed of 6.00 m/s is 12.5 m/s. Hence, this is the required solution.