28. Multiple-Concept Example 5 reviews many of the concepts that play roles in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands

7.78m/s

Explanation: As the the skier slide down the height, we assume the motion of a body, slidind down an incline plane.

Force down the plane= [email protected]

Frictional force= umg

u= coefficient of friction

Net force on skier = [email protected] umg

ma = [email protected]

a = g([email protected] - u) = 9.8 (sin 25- 0.2)

a = 9.8 × (0.4226-0.2) = 9.8 × 0.2226

a = 2.18m/s²

Using the formula V² = U² + 2aH

Where H = (10.4+ 3.5)=total height of descent before landing, U= 0.

V = √ 2 × 2.18× 13.9 = √60.604

V = 7.78m/s

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.