treegiraffe
Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.