Answer:
The athlete will rise 1.10 meters off the ground
Explanation:
Vertical Motion
If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by
[tex]\displaystyle y_m=\frac{v_o^2}{2g}[/tex]
The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at
[tex]\displaystyle a=\frac{F_n}{m}=\frac{2W}{m}=2g[/tex]
The speed at the end of his push can be computed by
[tex]v^2=2ay[/tex]
Replacing the value of a obtained above:
[tex]v^2=4gy[/tex]
where y is the length of this crouch
[tex]v^2=4\cdot 9.8\cdot 0.55[/tex]
[tex]v=4.64\ m/s[/tex]
This is the initial speed of this vertical launch, thus
[tex]\displaystyle y_m=\frac{4.64^2}{2\cdot 9.8}[/tex]
[tex]y_m=1.10\ m[/tex]