You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?

44.6 m

Explanation:

From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.

E = E'

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J

Substituting the values of the variables into the equation, we have

U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂

mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J

9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J

(735 kgm/s²)h + 75  kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J

(735 kgm/s²)h + 121.5  kgm²/s² = 29400 kgm²/s² + 3500 J

(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J

(735 kgm/s²)h + 121.5 J = 32900 J

(735 kgm/s²)h = 32900 J - 121.5 J

(735 kgm/s²)h = 32778.5 J

h = 32778.5 J/735 kgm/s²

h = 44.6 m

So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.

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