Question

A particle of mass m thrown vertically upward with velocity v0. The air resistance is mgcv² where c is a constant with v us the velocity at any time t. Show that the time taken by bthe particle to reach the highest point is given by v0 underroot c =tan(gt underroot c)

Answer 1

Upward motion

Using Newton’s 2nd law gives:

$$my''=-mgcy'^2-mg$$

$$\dfrac{dv}{dt}=-gcv^2-g$$

$$\int\dfrac{dv}{cv^2+1}=-\int g dt$$

$$\dfrac{1}{\sqrt{c}}\tan^{-1}(\sqrt{c}v)=-gt+C_1$$

$$t=0:C_1=\dfrac{1}{\sqrt{c}}\tan^{-1}(\sqrt{c}v_0)$$ $$\dfrac{1}{\sqrt{c}}\tan^{-1}(\sqrt{c}v)=-gt+\dfrac{1}{\sqrt{c}}\tan^{-1}(\sqrt{c}v_0)$$

When the particle reaches the highest point $v=0$

$$0=-gt_{vertex}+\dfrac{1}{\sqrt{c}}\tan^{-1}(\sqrt{c}v_0)$$

$$t_{vertex}=\dfrac{1}{g\sqrt{c}}\tan^{-1}(\sqrt{c}v_0)$$