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A skier waits at the top of a 11.7m hill. He then skis down the slope at an angle of 27 degrees above horizontal. What will his velocity be at the bottom of the hill? Show work

Respuesta :

Answer:

v = 3.89 m / s

Explanation:

For this exercise let's use the concept of conservation of mechanical energy

starting point. In the highest part of the mountain

         Em₀ = U = m g h

final point. In the lower part

         Em_f = K = ½ m v²

how energy is conserved

         Em₀ = Em_f

         mgh = ½ m v²

         v = [tex]\sqrt{ 2gh}[/tex]

let's use trigonometry to find the height

         sin θ = h / L

          h = L sin θ

we substitute

          v = [tex]\sqrt{ 2g L \ sin tea}[/tex]

let's calculate

          v = [tex]\sqrt{2 \ 9.8 \ 1.7 \ sin 27}[/tex]

          v = 3.89 m / s

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