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Tarzan, whose mass is 100 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 2.9 m above the ground and the bottom of his dangling feet are at a height 2.1 above the ground. When he first hits the ground he has dropped a distance 2.1, so his center of mass is (2.9-2.1) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.5 above the ground. Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?

Sagot :

boffeemadrid

Answer:

[tex]6.42\ \text{m/s}[/tex]

Explanation:

v = Final velocity

u = Initial velocity = 0

s = Displacement of Tarzan's feet = 2.1 m

a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

From the kinematic equations we get

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.1+0}\\\Rightarrow v=6.42\ \text{m/s}[/tex]

At the instant just before Tarzan's feet touch the ground the velocity of his feet are [tex]6.42\ \text{m/s}[/tex].

shubhamchouhanvrVT

The speed at an instant just before Tarzen's feet touch the ground will be  

[tex]V=6.42\ \dfrac{m}{s}[/tex]

What is speed?

The speed of any body is defined as the movement of the body with respect to the time.

It is given in the question that

Mass of the Tarzen m=100 kg

The distance of the Tarzen's feet from the ground s=2.1 m

The acceleration due to gravity [tex]a=9.81\ \frac{m}{s^2}[/tex]

The speed of the Tarzen will be calculated by applying the equation of motion.

[tex]v^2=u^2+2as[/tex]

Since initial velocity is zero u=0

[tex]v^2=2as[/tex]

[tex]v=\sqrt{2as}[/tex]

[tex]v=\sqrt{2\times 9.81\times 2.1} =6.42 \ \frac{m}{s}[/tex]

Thus the speed at an instant just before Tarzen's feet touch the ground will be  

[tex]V=6.42\ \dfrac{m}{s}[/tex]

To know more about speed follow

https://brainly.com/question/4931057

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