I'm completely stuck, I think I have to use Newton's second law but I have no idea where to start, any help would be appreciated!
At time $t=0$ a particle of unit mass is projected vertically upward with velocity $v_0$ against gravity, and the resistance of the air to the particle's motion is $k$ times its velocity. Show that during its flight the velocity $v$ of the particle at time $t$ is:
$$v = \left(v_{0} + \frac{g}{k}\right) e^{-kt} – \frac{g}{k}$$
Deduce that the particle reaches its greatest height when
$$t = \frac{1}{k} \ln\left({1+\frac{kv_{0}}{g}}\right)$$
and that the height reached is
$$ \frac{v_{0}}{k} – \frac{g}{k^2} \ln{\left(1 + \frac{kv_{0}}{g}\right)}$$
Thanks!
Best Answer
Start off by writing the equation:
$$\sum F = ma$$
$$ma=F_d +mg$$
where $F_d=kv$ is the drag force. and the particle is given to be of unit mass so the equation becomes.
$$-\frac{dv}{dt}=kv+g$$
$a=-\frac{dv}{dt}$ since velocity is decreasing,taking similar terms on same side and integrating with proper limits, we get
$$\int_{v_0}^v\frac{dv}{kv+g}=\int_o^t -dt$$
$$\frac{\ln(kv+g) - \ln(kv_0+g)}{k}=-t$$
$$\ln{\frac{kv+g}{kv_0+g}}=-kt$$
Solving for $v(t)$ gives us $$v = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$$
To find the time when max height is reached, put $v=0$, and find the corresponding value of $t$(say $t=t_0$)
$$(v_{0} + \frac{g}{k})e^{-kt_0}=\frac{g}{k}$$ this gives
$$t_0=\frac{1}{k} \ln\big({1+\frac{kv_{0}}{g}}\big)$$
To find max height
write $v = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$ as,
$$\frac{dx}{dt} = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$$ Integrating with proper limits, $$\int_0^h dx=\int_o^t (v_{0} + \frac{g}{k}) e^{-kt}.dt - \int_0^t\frac{g}{k}.dt$$
$$h=(v_0+\frac{g}{k})\frac{e^{-kt}}{-k} -(v_0+\frac{g}{k})\frac{1}{-k} -\frac{gt}{k}$$
putting $t=t_O$ in this equation to get max height,(since max height occurs when $v=0$, which happens when $t=t_0$)
$$ h_{max}=\frac{v_{0}}{k} - \frac{g}{k^2} \ln{(1 + \frac{kv_{0}}{g})}$$