According to Newton's second law:
"m\\frac{dv}{dt}=mg-kmv^2,\\\\\n\\space\\\\\n\\frac{dv}{dt}=g-kv^2,\\\\\n\\space\\\\\n\\frac{1}{g-kv^2}dv=dt,\\\\\n\\space\\\\\n\\int \\frac{1}{g-kv^2}dv=\\int dt,\\\\\n\\space\\\\\nt=\\frac{\\text{tanh}^{-1}\\bigg(v\\sqrt{k\/g}\\bigg)}{\\sqrt{gk}},"
"v=\\sqrt{\\frac{g}{k}}\\cdot\\text{tanh}\\big(t\\sqrt{gk}\\big)."
Just that simple. This equation describes how velocity depends with time.
But we have to find how the velocity depends on distance covered. Assume that the body is being acted upon by force of gravity "P". Use the same first equation:
"F_{net}=P-kmv^2,\\\\\n\\space\\\\\n\\frac{P}{g}v\\frac{dv}{dx}=P-kmv^2." Now put that
"a^2=\\frac{P}{km}=\\frac{g}{k}." Then
"v\\frac{dv}{dx}=g\\bigg(1-\\frac{v^2}{a^2}\\bigg),"
"-\\frac{vdv}{a^2-v^2}=-\\frac{g}{a^2}dx," integrate:
"\\text{ln}(a^2-v^2)=-2(g\/a^2)x+C." The initial conditions: "r_0=r_0, v_0=0".
"C=\\text{ln}a^2+\\frac{2gr_0}{a^2}." Therefore, the distance the lander falls in accelerating from zero velocity to "v" is:
"x=\\frac{a^2}{2g}\\bigg(\\text{ln}a^2+\\frac{2gr_0}{a^2}-\\text{ln}(a^2-v^2)\\bigg),\\\\\n\\space\\\\\nx=\\frac{a^2}{2g}\\bigg(\\text{ln}a^2-\\text{ln}(a^2-v^2)\\bigg)+r_0,\\\\\n\\space\\\\\nx=\\frac{1}{2k}\\bigg[\\text{ln}\\frac{g}{k}-\\text{ln}\\bigg(\\frac{g}{k}-v^2\\bigg)\\bigg]+r_0."
The calculations are quite difficult, so we can understand why that mistake at 340 mph happened.
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