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# Slings, Bullets, Blow-up, and Linearity

The sling is a simple weapon — essentially a pendulum with two strings instead of one, each of which holds a side of a cradle enclosing a projectile. When one spins the pendulum and releases a string, the discharged projectile can travel over $$400$$ meters. The sling was the world distance champion for millennia until the English long bow, and then the firearm, surpassed it. Speaking of the latter, a bullet shot into water has an unexpectedly short killing range despite its enormous speed.

Both of these phenomena—the surprisingly long killing distance of a sling and the surprisingly short killing distance of a water-bound bullet—involve near-infinite (on the human scale) velocities. Both can be “explained” by the fact that solutions of the simple ordinary differential equation (ODE) $$\dot x = x ^2$$ blow up in finite time, something usually presented as a mere scholastic curiosity.

### Launching the Sling

Figure 1. The velocity v blows up in finite time T=the tension of the string, giving rise to the accelerations atangential and acentripetal.
The caricature of the sling in Figure 1 shows the hand moving in a circle so as to make the cradle describe a concentric circle (to achieve this, the hand must accelerate a certain way). I claim that the speed $$v$$ then satisfies the ODE mentioned above: $$\dot v = k v ^2$$, with a constant $$k$$. Indeed, Figure 1 gives us $$a_{\rm tangential}=a_{centripetal} \tan \theta$$; and since $$a_{\rm tangential} = \dot v$$ and $$a_{centripetal} = v ^2 /R$$, this yields

$\dot v = k v^2, \:\:\:\:\:\: k = \tan \theta /R.$

And the solution

$v= \frac{v_0}{1 - v_0k t }$

does in fact approach infinity as $$t \rightarrow t_\infty = 1/v_0k$$ in the idealized non-relativistic world of infinitely fast hands and infinitely strong strings. The “shadow” of this infinite speed is seen in the fact that a good slinger can launch at over $$1/5$$th the speed of sound (computed from the aforementioned $$>400$$ m distance. In fact, the actual speed is higher, given that the calculation ignores the air resistance).

### Shooting into the Water

Figure 2. Shooting into water.
Continuing with the weapons-related theme, let us ask: what depth renders harmless a bullet shot down into the water? Assume that (i) the water drag on the bullet is proportional to the square of the speed; (ii) the terminal sinking velocity of the bullet is $$1$$ m/sec, and (iii) the safe velocity of the bullet is $$v_{\rm s}= 10$$ m/sec (the speed gained in dropping $$15$$ feet, painful but probably not dangerous). With these assumptions, the safe depth turns out to be

$x_s \approx 0.1\ln \biggl( 1+ \frac{v_0}{v_s} \biggl), \tag1$

in meters. Before deriving this expression, let us find safe depths for various bullet speeds. For $$v_0= 10^3$$ m/s (about three times the speed of sound), $$x_s\approx 46$$ cm. For the escape velocity $$v_0\approx 11$$ km/sec, $$x_s\approx 69$$ cm. Every increase in the order of magnitude simply adds about $$23$$ cm to the safety depth. And for a bullet travelling at the speed of light (here I am abandoning the last touch with reality), the safe depth is just $$2$$ m! This drastic loss of speed of the bullet is identical, up to the time-reversal, to the sling projectile’s drastic gain of speed.

### Derivation of (1)

The bullet shot straight down obeys Newton’s law: $$m \dot v = - c v ^2 + mg$$, or

$\dot v=-kv^2+g, \:\:\:\:\:\: k=\frac{c}{m}, \tag2$

where $$v = \dot x$$ and the $$x$$-axis points down. From the terminal velocity condition $$- k v_{\rm term} ^2 + g = 0$$, we find $$k = g/ v_{\rm term}^2\approx 10 m ^{-1}$$. Neglecting $$g$$ in $$(2)$$, we get

$\dot v = - k v ^2, \:\:\:\:\:\: k \approx 10m ^{-1}.$

Substituting the initial condition $$v(0)=v_0$$ and the time $$t_s$$ of reaching velocity $$v_s$$ into the solution of this ODE gives

$v_s= \frac{v_0}{1+ kv_0\,t_s} \Rightarrow t_s= k ^{-1} (v_s ^{-1} - v_0 ^{-1} )< \frac{1}{kv_s}.$

Integrating $$v$$ gives $$x= k ^{-1} \ln (1+ kv_0t)$$, and substituting $$t_s<1/kv_s$$ into this expression gives $$(1)$$.

### The Hidden Linearity

Figure 3. Geometry of the finite-time blow-up. The slope x of the line of sight satisfies the “sling” ODE.
The ODE $$\dot x = x ^2$$ (a special case of the Riccati equation) hides linear growth, which can be expressed in two equivalent ways. Algebraically, the equation simply amounts to the linear growth of $$1/x$$, namely to $$\frac{d}{dt} \frac{1}{x} = 1$$. Geometrically, this ODE governs the evolution of the slope $$x=v/u$$ of solution vectors in the $$(u,v)$$-plane of the trivial system $$\dot u = -1, \ \dot v = 0$$. Figure 3 offers an essentially equivalent realization. The blow-up of the solution occurs when the plane is overhead.

As a concluding remark, the assumption $$dv/dt=-kv^2$$ amounts to a more natural-sounding statement: the kinetic energy $$E$$ decays exponentially with the distance $$dE/dx=-2kE$$.