Question

What is the terminal speed for a 79 kg skier going down a 37 ∘ snow-covered slope on wooden skis μk= 0.060? Assume that the skier is 1.8 m tall and 0.50 m wide. Assume the skier's drag coefficient is 0.80.

Answer 1

The free body diagram can be drawn as,

  

drag force frictionNiarce mgsin mgcos

Now the force equation can be written as,

  $\small mgsin\theta-friction \: force-drag\: force=0$

  $\small \Rightarrow mgsin\theta-\mu mgcos\theta -\frac{1}{2}\rho v_{T}^{2}C_{d}A=0$

Here _{$\small \rho$} = air density = 1.29 kg/m³ , C_d = drag coefficient = 0.80, A = area of skier = (1.8)(0.50) = 0.9 m². Thus the terminal velocity v_T can be calculated as,

$\small \Rightarrow mgsin\theta-\mu mgcos\theta =\frac{1}{2}\rho v_{T}^{2}C_{d}A$

$\small \Rightarrow v_{T}=\sqrt{\frac{2mg(sin\theta -\mu cos\theta )}{\rho C_{d}A}}$

$\small \Rightarrow v_{T}=\sqrt{\frac{2(79)(9.8)(sin37^{\circ} -(0.060) cos37^{\circ})}{1.29(0.80)(0.9)}}$

$\small \Rightarrow v_{_{_{T}}}=30.4\: m/s$