Roller Coaster Physics: Calculating Velocity at 18 Meters After a 22-Meter Drop

A 236 kg roller coaster car released from rest at 40 m height reaches 18 m height. Using conservation of energy (ignoring friction):

Height change: Δy = 18 m − 40 m = −22 m

Potential energy loss: ΔU = m g Δy = (236 kg)(9.8 m/s²)(22 m) = 50,906 J

This converts to kinetic energy: ΔK = ½ m v²

½ (236 kg) v² = 50,906 J → v² = 431.4 → v ≈ 20.8 m/s

Correction: The original calculation used g = 9.8 and Δy = -22, yielding v = 28.2 m/s. Recalculating: v = √(2 × 9.8 × 22) = √431.2 ≈ 20.8 m/s. The earlier value (28.2 m/s) appears to have used an incorrect energy difference. The correct velocity at 18 m is approximately 20.8 m/s (75 km/h).