Answer to Question #200468 in Math for Raghav

Question

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

2gx/V_t² = ln[(u_o²+V_t²)/(V²+V_t²)], where k=g/V_t²

Answer 1

Using Newton’s 2nd law gives:

$$m\dfrac{d^2x}{dt^2}=-mk(\dfrac{dx}{dt})^2-mg$$

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

$$\dfrac{dV}{dt}=\dfrac{dV}{dx}\cdot\dfrac{dx}{dt}=V\dfrac{dV}{dx}$$

$$V\dfrac{dV}{dx}=-kV^2-g$$

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

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

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

$z=kV^2+g, dz=2kVdV$

$VdV=\dfrac{1}{2}dz$

$$\int \dfrac{VdV}{kV^2+g}=\int \dfrac{dz}{2kz}=\dfrac{1}{2k}\ln|z|+C_1$$

$$=\dfrac{1}{2k}\ln(kV^2+g)+C_1$$

$$\dfrac{1}{2k}\ln(kV^2+g)=-x+\dfrac{1}{2}C_2$$

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

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

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

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

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

$$\dfrac{2gx}{V_t^2}=\ln(\dfrac{u_0^2+V_t^2}{V^2+V_t^2})$$