Answer to Question #200468 in Math for Raghav

Question #200468

Consider the upward motion of a particle under gravity with a velocity of projection uo and resistance mkv2. Show that the velocity V at the time t and distance x from point lf projection are related as

2gx/Vt2 = ln[(uo2+Vt2)/(V2+Vt2)], where k=g/Vt2

1
Expert's answer
2021-06-07T15:33:34-0400

Using Newton’s 2nd law gives:


md2xdt2=mk(dxdt)2mgm\dfrac{d^2x}{dt^2}=-mk(\dfrac{dx}{dt})^2-mg

dVdt=kV2g\dfrac{dV}{dt}=-kV^2-g

dVdt=dVdxdxdt=VdVdx\dfrac{dV}{dt}=\dfrac{dV}{dx}\cdot\dfrac{dx}{dt}=V\dfrac{dV}{dx}

VdVdx=kV2gV\dfrac{dV}{dx}=-kV^2-g

VdVkV2+g=dx\dfrac{VdV}{kV^2+g}=-dx

VdVkV2+g=dx\int \dfrac{VdV}{kV^2+g}=-\int dx

VdVkV2+g\int \dfrac{VdV}{kV^2+g}

z=kV2+g,dz=2kVdVz=kV^2+g, dz=2kVdV


VdV=12dzVdV=\dfrac{1}{2}dz


VdVkV2+g=dz2kz=12klnz+C1\int \dfrac{VdV}{kV^2+g}=\int \dfrac{dz}{2kz}=\dfrac{1}{2k}\ln|z|+C_1


=12kln(kV2+g)+C1=\dfrac{1}{2k}\ln(kV^2+g)+C_1


12kln(kV2+g)=x+12C2\dfrac{1}{2k}\ln(kV^2+g)=-x+\dfrac{1}{2}C_2

2kx=ln(kV2+g)+kC22kx=-\ln(kV^2+g)+kC_2

t=0:0=ln(ku02+g)+kC2t=0: 0=-\ln(ku_0^2+g)+kC_2

=>kC2=ln(ku02+g)=>kC_2=\ln(ku_0^2+g)

2kx=ln(ku02+gkV2+g)2kx=\ln(\dfrac{ku_0^2+g}{kV^2+g})

If k=gVt2k=\dfrac{g}{V_t^2}


2gxVt2=ln(u02+Vt2V2+Vt2)\dfrac{2gx}{V_t^2}=\ln(\dfrac{u_0^2+V_t^2}{V^2+V_t^2})



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