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599 N skier begins on a hill 32 m above the valley floor. He travels down to the valley floor and up a 18 m hill on the other side. Ignoring air resistance, what is his velocity at the top of the 18 m hill?

Work please!

Sagot :

Answer:

v = 129.5 m / s

Explanation:

For this exercise we must use the conservation of mechanical energy.

Starting point. On the side of the first hill

           Em₀ = U = m g h₁

Final point. On the other hill

           Em_f = K + U = ½ m v² + mgh₂

where h2 is the latura of the ora hill

as they indicate that there is no friction, energy is conserved

            Em₀ = Em_f

             mgh₁ = ½ mv² + m gh₂

            v² = 2 mg (h1-h2)

            v = [tex]\sqrt{2m g( h_1 - h_2)}[/tex]

let's calculate

           v = [tex]\sqrt{ 2 \ 599 \ ( 32-18)}[/tex]

           v = 129.5 m / s

This speed is horizontal at the top of the hill

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