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?

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:

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

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s