Question

show that in the motion of a heavy particle in a medium the resistance of which varies as the velocity, the greatest height abov the level of the point of projection is reached in less than half the total time of flight above that level.

Answer 1

Let $v_0\&\theta$ is the initial velocity and angle of the projectile respectively and for continence choose mass $m=1kg$ .

As, the particle moves in the restive medium, thus the net force acting on the particle is ,

$$F=ma_y=-kv_y-mg\implies a_y=-kv_y-g \: (\square)$$

Thus,

$$\frac{dv_y}{dt}=-kv_y-g\\ \implies \frac{dv_y}{kv_y+g}=-dt\\ \implies \int \frac{dv_y}{kv_y+g}=\int-dt\\ \implies \frac{ ln(kv_y+g)}{k}=-t+c$$

since, at $t=0,v_y=u=v_0\sin(\theta)$ , which implies

$$c=\frac{ ln(kv_0\sin(\theta)+g)}{k}\\ \implies ln\bigg(\frac{kv_y+g}{kv_0\sin(\theta)+g}\bigg)=-kt\\ \implies v_y=\frac{(kv_0\sin(\theta)+g)e^{-kt} -g}{k} \hspace{1cm}(\clubs)$$

When $v_y=0,t=T_a$ ,particle reaches maximum height(time of ascent), thus from $(\clubs)$ we get,

$$\frac{(kv_0\sin(\theta)+g)e^{-kT_a} -g}{k}=0\\ \implies T_a=-\frac{1}{k}ln\bigg(\frac{g}{kv_0\sin(\theta)+g}\bigg)\hspace{1cm}(\spades)$$

Now, we will calculate the height$(H)$ at $t=T_a$ ,thus from $(\clubs)$ we get,

$$\frac{dy}{dt}=\frac{(kv_0\sin(\theta)+g)e^{-kt} -g}{k} \\ \implies \int_{0}^{H} dy =\int_{0}^{T_a} \frac{(kv_0\sin(\theta)+g)e^{-kt} -g}{k} dt\\ \implies \frac{(kv_0\sin(\theta)+g)e^{-kt} }{-k^2}-\frac{gt}{k} \bigg|_{0}^{T_a}=H\\ \implies \bigg(\frac{(kv_0\sin(\theta)+g)e^{-kT_a} }{-k^2}-\frac{gT_a}{k}\bigg)-\\\bigg( \frac{kv_0\sin(\theta)+g }{-k^2}\bigg)=H\\ \implies \frac{u }{k} -\frac{g}{k}T_a=H\hspace{1cm}(\star)\\$$

Now, will calculate time of descent $T_d$ . Clearly, the acceleration of the particle will be

$$a_y=g-kv_y \\ \implies \frac{dv_y}{dt}=g-kv_y\\ \implies \frac{dv_y}{g-kv_y}=dt\\ \implies\frac{1}{k} ln(\frac{kv_y-g}{g})=-t\\ \implies v_y=\frac{g}{k}(1-e^{-kt}) \hspace{1cm}(\star \star)$$

Using the above result and $(\star)$ and after further calculation we will get,

$$H=\frac{g}{k} \int_{0}^{T_d} (1-e^{-kt})dt\\ \implies \frac{u }{k} -\frac{g}{k}T_a=\frac{g}{k} (T_d+\frac{e^{-kT_d}}{k} -\frac{1}{k} )\\ \implies T_a +T_d=T_{flight}=\frac{u}{g} +\frac{1}{k} -\frac{e^{-kT_d}}{k}\\ \implies T_{flight}-2T_a=\frac{u}{g} +\frac{1}{k} -\frac{e^{-kT_d}}{k}-\\\frac{2}{k}ln(1+\frac{ku}{g})\hspace{1cm}(\dag)$$

Now, expanding up to second order we get,

$$ln(1+\frac{ku}{g})=\frac{ku}{g}-\frac{1}{2}(\frac{ku}{g})^2$$

On plugin this result to $(\dag)$ we get,

$$T_{flight}-2T_a=\frac{1}{k}(1-e^{-kT_d})+\frac{u}{g^2}(ku-g)\\ \implies T_{flight}-2T_a >0 \hspace{1cm}(\because 1-e^{-kT_d}>0\&ku-g\geq0)\\ \implies T_a<\frac{1}{2}T_{flight}$$

Alternatively, from physical point of view, we can clearly see from $(\square)$ and$(\star \star)$ that the rate of velocity more rapidly negative in $(\star \star)$ than $(\square)$ ,therefore it is obvious that $T_a<T_d$ ,Hence the result.