Given:
Mass of skier = 65.0 kg
Length of Slope, h = 42.0 m
Angle of slope = 37.2°
Let's find how fast she is going at the bottom.
To find the speed at the bottom, we have:
[tex]\begin{gathered} \Delta x=\frac{h}{\sin \theta} \\ \\ \Delta x=\frac{42.0}{\sin 37.2} \\ \\ \Delta x=\frac{42.0}{0.605} \\ \\ \Delta x=64.5m \end{gathered}[/tex]
Now, to find the speed apply the fomula:
[tex](v_x)^2_f=(v_x)_i^2+2a_x\Delta x[/tex][tex]\begin{gathered} \text{Where:} \\ a_x=\frac{\Sigma fx_{}}{m}=\frac{(w\sin \theta)-\mu_k(w\cos \theta)}{m}=\frac{(65\ast9.8\sin37.2)-(0.107(65\ast9.8\cos37.2))}{65} \\ \\ a_x=\frac{330.8389535}{65} \\ \\ a_x=5.089 \end{gathered}[/tex]
Thus, we have:
[tex]\begin{gathered} (v_x)^2_f=0+2(5.089)(42)=427.54 \\ \\ (v_x)_f=\sqrt[]{427.54}=20.7\text{ m/s} \end{gathered}[/tex]
Therefore, the skiers speed from the bottom is 20.7 m/s
ANSWER:
20.7 m/s
