Question

a skier starts from rest at the top of a 45.0 m hill, coasts down a 30° slope into a valley and continues up to the top of a 40.0 m hill. both hill heights are measured from the valley floor. assume the skier puts no effort into the motion (they always coast) and there is no friction. how fast will the skier be moving on the valley floor between the two hills? a 29.7 m/s b 24.8 m/s c 12.4 m/s d 37.8 m/s

Explanation:

Step1: Apply conservation of mechanical energy

The initial state is at the top of the \(45.0\) - m hill with \(v_{i}=0\) (rest), so the initial mechanical energy \(E_{i}=mgh_{1}\), where \(h_{1} = 45.0\) m. The state at the valley floor has mechanical energy \(E = mgh_{0}+\frac{1}{2}mv^{2}\), where \(h_{0} = 0\) (reference level). By conservation of energy \(E_{i}=E\), so \(mgh_{1}=mgh_{0}+\frac{1}{2}mv^{2}\). Since \(h_{0} = 0\), the equation simplifies to \(gh_{1}=\frac{1}{2}v^{2}\).

Step2: Solve for \(v\)

We know that \(g = 9.8\space m/s^{2}\) and \(h_{1}=45.0\space m\). Substituting into \(v=\sqrt{2gh_{1}}\), we get \(v=\sqrt{2\times9.8\times45.0}\).

$$
LATEXBLOCK0
$$

Answer:

A. \(29.7\space m/s\)