Projectile motion with air resistance. A projectile S of mass m is fired from a gun with muzzle speed Vo at an angle β with the horizontal plane. Neglect the Earth’s motion and wind effects and assume that air resistance is govern ed by Stokes’s law. Determine the projectile’s motion as a function

Question

Projectile motion with air resistance. A projectile S of mass m is fired from a gun with muzzle speed [latex]v_0[/latex] at an angle β with the horizontal plane. Neglect the Earth’s motion and wind effects and assume that air resistance is govern ed by Stokes’s law. Determine the projectile’s motion as a function of time.

Accepted Answer

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 [latex]\mathbf{v}_0=v_0(\cos \beta \mathbf{i} + \sin \beta \mathbf{j})[/latex] yields

[latex]\ddot{x}=-v \dot{x}, \quad \ddot{y}=-g - v \dot{y} \quad ext { with } \quad v \equiv \frac{c}{m} .[/latex] (6.36)

[latex]\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 .[/latex]

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

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

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

[latex]x(t)=\frac{v_0 \cos \beta}{v}\left(1 - e^{-v t} ight)[/latex] (6.37b)

[latex]y(t)=-\frac{g}{v} t + \frac{1}{v}\left(v_0 \sin \beta + \frac{g}{v} ight)\left(1 - e^{-v t} ight)[/latex] (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 [latex]t ightarrow \infty[/latex], in the absence of impact, (6.37a) gives [latex]\dot{x} ightarrow 0[/latex] and [latex]\dot{y} ightarrow-g / u[/latex]. Hence, the projectile attains the terminal speed [latex]v_{\infty}=g / u[/latex] 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 [latex]r_{\infty} \equiv \lim _{t ightarrow \infty} x(t)=\left(v_0 \cos \beta ight) / v[/latex] 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 .

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

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