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A 75 kg skier accelerates from rest to a speed of 12 m/s in 3.0 seconds on a frictionless hill. What is the slope of the hill?

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Onyebuchinnajiidn

The  slope of the hill on which the skier is accelerating is 24⁰.

The given parameters;

  • mass of the skier, m = 75 kg
  • final velocity, v = 12 m/s
  • time of motion., t = 3 s

The acceleration  of the skier is calculated as follows;

[tex]a = \frac{v}{t} \\\\a = \frac{12}{3} = 4 \ ms^2[/tex]

The net force on the skier is calculated as;

[tex]mgsin\theta - \mu_kF_n = ma\\\\mgsin\theta- 0 = ma\\\\gsin\theta = a\\\\sin\theta = \frac{a}{g} \\\\\theta = sin^{-1} (\frac{a}{g} )\\\\\theta = sin^{-1} (\frac{4}{9.8} )\\\\\theta = 24 \ ^0[/tex]

Thus, the  slope of the hill on which the skier is accelerating is 24⁰.

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