Answered

Tarzan, whose mass is 96 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.1 m above the ground and the bottom of his dangling feet are at a height 1.3 above the ground. When he first hits the ground he has dropped a distance 1.3, so his center of mass is (2.1 - 1.3) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.4 above the ground.

Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?

Answer :

MechEngineer

Answer:

5.05 m/s

Explanation:

The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:

[tex]mgH = mgh + mv^2/2[/tex]

Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m

[tex]gH = gh + v^2/2[/tex]

[tex]v^2 = 2g(H - h)[/tex]

[tex]v^2 = 2*9.81(2.1 - 0.8) = 25.506 [/tex]

[tex]v = \sqrt{25.506} = 5.05 m/s[/tex]

Other Questions