Answer:
She raised approximately 4 meters as she crosses the bar
Explanation:
Energy Conservation
In the absence of air resistance and any other energy-absorbing events, the total mechanical energy is conserved. Recall the mechanical energy is the sum of the gravitational potential energy and the kinetic energy:
[tex]\displaystyle M=m.g.h+\frac{1}{2}m.v^2[/tex]
When our athlete is at the ground level her height is 0, thus she has only kinetic energy given by the speed she's at when starting the jump:
[tex]\displaystyle M_1=m.g.(0)+\frac{1}{2}m.v^2=\frac{1}{2}m.v^2[/tex]
Later when she's directly above the bar, she has both energies since some speed remains at a certain height h we must calculate. The mechanical energy is
[tex]\displaystyle M_2=m.g.h+\frac{1}{2}m.v'^2[/tex]
Equating both energies
[tex]\displaystyle m.g.h+\frac{1}{2}m.v'^2=\frac{1}{2}m.v^2[/tex]
Simplifying by m and rearranging
[tex]\displaystyle m.g.h=\frac{1}{2}m.v^2-\frac{1}{2}m.v'^2=\frac{1}{2}(v^2-v'^2)[/tex]
Solving for h
[tex]\displaystyle h=\frac{(v^2-v'^2)}{2g}=\frac{(9.2^2-2.4^2)}{2\cdot 9.8}[/tex]
[tex]\boxed{h=4.02\ m}[/tex]
She raised approximately 4 meters as she crosses the bar