Question

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is she going when she reaches the bottom?

Answers

  1. Answer:

    v = 45.37 m/s

    Explanation:

    Given,

    angle of inclination = 8.0°

    Vertical height, H  = 105 m

    Initial K.E. = 0 J

    Initial P.E. = m g H

    Final PE = 0 J

    Final KE = \dfrac{1}{2}mv^2

    Using Conservation of energy

    KE_i + PE_i + KE_f + PE_f

    0 + m g H = \dfrac{1}{2}mv^2 + 0

    v = \sqrt{2gH}

    v = \sqrt{2\times 9.8 \times 105}

    v = 45.37 m/s

    Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

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