Answer:
She raised approximately 4 meters as she crosses the bar
Step-by-step 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:
![\displaystyle M=m.g.h+(1)/(2)m.v^2](https://img.qammunity.org/2021/formulas/physics/college/3khcumoiekl9p2s833nrzmgc3zrtzysqkf.png)
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:
![\displaystyle M_1=m.g.(0)+(1)/(2)m.v^2=(1)/(2)m.v^2](https://img.qammunity.org/2021/formulas/physics/college/objlfen7t4v9b94c4lmbr2zzmzhbo9x10r.png)
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
![\displaystyle M_2=m.g.h+(1)/(2)m.v'^2](https://img.qammunity.org/2021/formulas/physics/college/pfb3ars6g39f877uvje88ebe3q264239wz.png)
Equating both energies
![\displaystyle m.g.h+(1)/(2)m.v'^2=(1)/(2)m.v^2](https://img.qammunity.org/2021/formulas/physics/college/bh45d3hsns7854kv6q6mvjtyl7em8vr1ez.png)
Simplifying by m and rearranging
![\displaystyle m.g.h=(1)/(2)m.v^2-(1)/(2)m.v'^2=(1)/(2)(v^2-v'^2)](https://img.qammunity.org/2021/formulas/physics/college/ms9mk606dfhmqvrw5v4ziaiwl28h8enaj5.png)
Solving for h
![\displaystyle h=((v^2-v'^2))/(2g)=((9.2^2-2.4^2))/(2\cdot 9.8)](https://img.qammunity.org/2021/formulas/physics/college/vn5fdcj189m64gh4nfmiorxxuia18lb8n9.png)
![\boxed{h=4.02\ m}](https://img.qammunity.org/2021/formulas/physics/college/8ft2wghhob3eh13qcwg755a3dobdiwpkpf.png)
She raised approximately 4 meters as she crosses the bar