1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

[tex]\frac{1}{2}mv^2 = mgh[/tex]

where

m is the mass of the man

v is the speed after jumping

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

h = 2.0 – 1.0 = 1.0 m is the change in height

Solving for v,

[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s[/tex]

In the acceleration phase, we know that the initial velocity is

[tex]u=0[/tex]

And the force exerted on the floor is 2.3 times the gravitational force, so

[tex]F=2.3 mg[/tex]

This means the net force on you is

[tex]F_{net} = F-mg=2.3mg-mg=1.3 mg[/tex]

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

[tex]a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g[/tex]

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

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

where s is the quantity we want to find. Solving for s,

[tex]s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m[/tex]

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

[tex]\Delta h = -0.77 m[/tex]

This means that the lowest point reached by the center of mass above the floor during the crouch is

[tex]h’=h+\Delta h = 1.0 – 0.77 = 0.23 m[/tex]

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

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