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.
