A body is dropped from rest in a medium with quadratic resistance γv². The terminal speed is V=√(g/γ). The distance fallen by the time its speed becomes V/2 is:
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A body is dropped from rest in a medium with quadratic resistance γv². The terminal speed is V=√(g/γ). The distance fallen by the time its speed becomes V/2 is:

G1 Physics · Accepted Answer

Take downward as positive. The acceleration is a = g−γv² = g[1−v²/V²]. Use a = v dv/dx: v dv/dx = g[1−v²/V²]. Thus dx = v dv / {g[1−v²/V²]}. Integrate from v=0 to V/2: x = −(V²/2g)ln[1−v²/V²]₀→V/2 = −(1/2γ)ln(3/4) = ln(4/3)/(2γ).