A particle is projected vertically upward with speed u in a medium with quadratic air resistance. Let V=√(g/γ) be the terminal falling speed and q=u/V. The particle returns to its point of projection. Its total time of flight and impact speed are respectively:
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Question

A particle is projected vertically upward with speed u in a medium with quadratic air resistance. Let V=√(g/γ) be the terminal falling speed and q=u/V. The particle returns to its point of projection. Its total time of flight and impact speed are respectively:

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G1 Physics · Accepted Answer

Ascent time is ascent time = (V/g)tan⁻¹q. Maximum height is H = (V²/2g)ln(1+q²). During descent, take downward speed w as positive: dw/dt = g[1−w²/V²]. Using w dw/dx, H = (V²/2g)ln[1/(1−wᵢ²/V²)]. Equating the two expressions for H: 1/(1−wᵢ²/V²)=1+q². Hence wᵢ = Vq/√(1+q²) = u/√(1+q²). Descent time from rest to wᵢ is descent time = (V/g)tanh⁻¹(wᵢ/V). Since tanh⁻¹[q/√(1+q²)] = sinh⁻¹q, descent time = (V/g)sinh⁻¹q. Therefore T = (V/g)[tan⁻¹q+sinh⁻¹q], and impact speed is u/√(1+q²).