A body of mass $m$ is projected vertically upwards with velocity $v_0$. If it experiences a resistive force proportional to its velocity, $F=-kv$, what is the maximum height reached by the body?

Solution

1. The equation of motion for the body is $m\frac{dv}{dt}=-mg-kv$.
2. Using the chain rule $\frac{dv}{dt}=v\frac{dv}{dy}$, we rewrite the equation as $mv\frac{dv}{dy}=-(mg+kv)$.
3. Rearranging and integrating from the initial velocity $v_0$ at $y=0$ to $v=0$ at $y=H$ gives ${\int }_0^{H}dy={\int }_{v_0}^0\frac{mv}{-mg-kv}dv$.
4. Evaluating the integral yields $$\begin{array}{rl}H& =\frac{m}{k}{\int }_0^{v_0}\frac{v}{v+\frac{mg}{k}}dv\\ & =\frac{m}{k}{\left[v-\frac{mg}{k}\ln (1+\frac{kv}{mg})\right]}_0^{v_0}\end{array}$$.
5. Simplifying the expression results in $H=\frac{m{v}_0}{k}-\frac{m^{2}g}{k^2}\ln (1+\frac{k{v}_0}{mg})$.
6. Hence the answer is (B).