Question 6.12: Projectile motion with air resistance. A projectile S of mas...

The equations of motion with air resistance governed by Stokes’s law are given in (6.36) . To find the motion x(S, t) , we first integrate the system (6.36) to obtain v(S, t ). Use of the initial condition  \mathbf{v}_0=v_0(\cos \beta \mathbf{i}  +  \sin \beta \mathbf{j})  yields

\ddot{x}=-v \dot{x}, \quad \ddot{y}=-g  –  v \dot{y} \quad \text { with } \quad v \equiv \frac{c}{m} .              (6.36)

\int_{v_0 \cos \beta}^{\dot{x}} \frac{d \dot{x}}{\dot{x}}=-v t, \quad \int_{v_0 \sin \beta}^{\dot{y}} \frac{d \dot{y}}{g  +  v \dot{y}}=-t .

These deliver the projectile’s velocity components as functions of time:

\dot{x}=\left(v_0 \cos \beta\right) e^{-v t}, \quad \dot{y}=-\frac{g}{v}  +  \left(v_0 \sin \beta  +  \frac{g}{v}\right) e^{-v t}.          (6.37a)

Then integration of (6.37a) with the initial condition  \mathbf{x}_0=\mathbf{0}  yields the motion of the projectile as a function of time:

x(t)=\frac{v_0 \cos \beta}{v}\left(1  –  e^{-v t}\right)                 (6.37b)

y(t)=-\frac{g}{v} t  +  \frac{1}{v}\left(v_0 \sin \beta  +  \frac{g}{v}\right)\left(1  –  e^{-v t}\right)             (6.37c)

Let us imagine that the projectile is fired from a hilltop into a wide ravine, as shown in Fig. 6.10. Then, as  t \rightarrow \infty,  in the absence of impact, (6.37a) gives  \dot{x} \rightarrow 0  and  \dot{y} \rightarrow-g / \nu.  Hence, the projectile attains the terminal speed  v_{\infty}=g / \nu   at which its weight is balanced by air  resistance; and (6.37b) and (6.37c) show that the projectile approaches asymptotically, the vertical range line at  r_{\infty} \equiv \lim _{t \rightarrow \infty} x(t)=\left(v_0 \cos \beta\right) / v  in Fig. 6.10. In the absence of air resistance, the range for the same situation would grow indefinitely with the width of the ravine. The simple Stokes model thus provides a more realistic picture of projectile motion with air resistance that limits its range .