Given:
The initial height of the rock was,
![h_i=5\text{ m}](https://img.qammunity.org/2023/formulas/physics/college/41lk179wp2ftmh5nh48fl7eejsimwe87kr.png)
The final height of the rock is,
![h_f=1.5\text{ m}](https://img.qammunity.org/2023/formulas/physics/college/3bqoi7fc4eeeedlb1an8mr4nm03pp7qjdb.png)
The mass of the rock is,
![m=0.5\text{ kg}](https://img.qammunity.org/2023/formulas/physics/college/555ko62gewv39qrrumgc3ocstrzolksq8n.png)
The travelling speed of the rock is,
![v^(\prime)=8.0\text{ m/s}](https://img.qammunity.org/2023/formulas/physics/college/tlzgd3rdb2v3pnyvy382gkxxv8mg4h3im7.png)
To find:
a) The speed of the rock at Joe's head
b) how much work was done by air resistance
Step-by-step explanation:
The displacement of the rock is,
![\begin{gathered} h=h_i-h_f \\ =5-1.5 \\ =3.5\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/s64gfq6ie9ie2eed8qm6y6s6p2024wemks.png)
The final speed at Joe's head is,
![\begin{gathered} v=√(u^2+2gh) \\ =√(0^2+2*9.8*3.5) \\ =√(68.6) \\ =8.28\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/yx1fak9i23bqzxsfa83sot6dnjv3anqpn2.png)
Hence, the speed of the rock at Joe's head is 8.28 m/s.
b)
The speed at Joe's head was 8.0 m/s, and the loss of kinetic energy is,
![\begin{gathered} (1)/(2)* m[(8.23)^2-(8.0)^2] \\ =(1)/(2)*0.5*[3.73] \\ =0.93\text{ J} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/sb976amd8c5yimrj8t5eruyr70fcl2kcuv.png)
This loss of energy is the work done by the air resistance.
Hence, the work done by the air resistance is 0.93 J.