Given:
The initial height of the rock was,
[tex]h_i=5\text{ m}[/tex]The final height of the rock is,
[tex]h_f=1.5\text{ m}[/tex]The mass of the rock is,
[tex]m=0.5\text{ kg}[/tex]The travelling speed of the rock is,
[tex]v^{\prime}=8.0\text{ m/s}[/tex]To find:
a) The speed of the rock at Joe's head
b) how much work was done by air resistance
Explanation:
The displacement of the rock is,
[tex]\begin{gathered} h=h_i-h_f \\ =5-1.5 \\ =3.5\text{ m} \end{gathered}[/tex]The final speed at Joe's head is,
[tex]\begin{gathered} v=\sqrt{u^2+2gh} \\ =\sqrt{0^2+2\times9.8\times3.5} \\ =\sqrt{68.6} \\ =8.28\text{ m/s} \end{gathered}[/tex]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,
[tex]\begin{gathered} \frac{1}{2}\times m[(8.23)^2-(8.0)^2] \\ =\frac{1}{2}\times0.5\times[3.73] \\ =0.93\text{ J} \end{gathered}[/tex]This loss of energy is the work done by the air resistance.
Hence, the work done by the air resistance is 0.93 J.
