A projectile problem

Here’s a projectile problem: at what angle should you throw a stone such that it is always travelling away from you?

If you throw a stone towards the horizon, it most likely won’t come back to you (unless you throw it at someone). If you throw a stone straight upwards, gravity will pull it back towards you. Then surely there exists an angle below which the stone never gets any closer to you?

To clarify, we are measuring the launch angle \theta relative to the ground. Assuming the stone is launched from the origin at time t=0, the equations describing its coordinates are well known to be

x(t) = v\cos\theta\ t

y(t) = v \sin\theta\ t - \frac{1}{2}gt^2

where v is the speed at which you throw the stone and g is the magnitude of acceleration due to gravity.

The distance of the stone from you (the origin) is then

r=\sqrt{x^2+y^2}

The question posed mathematically reads: what are the constraints on \theta such that

\displaystyle \frac{dr(t)}{dt}>0\ \forall\ t\ge 0

It is slightly neater to instead to work with the distance squared:

\displaystyle \frac{dr^2(t)}{dt}> 0

It does no harm to do this since, if the distance squared is increasing, so too is the distance.

Explicitly,

r^2=x^2+y^2

r^2=v^2\cos^2\theta\ t^2+ (v \sin\theta\ t - \frac{1}{2}gt^2)^2

which on rearrangement gives

r^2=v^2 t^2-gv\sin\theta\ t^3+\frac{1}{4}g^2t^4

Differentiating this gives

\displaystyle \frac{dr^2}{dt}=2v^2 t-3gv\sin\theta\ t^2+g^2t^3

\displaystyle \frac{dr^2}{dt}=t(2v^2-3gv\sin\theta\ t+g^2t^2)

What are the conditions under which this function is always greater than or equal to 0? Since t\ge 0, we must demand that the contents of the brackets satisfy

2v^2-3gv\sin\theta\ t+g^2t^2> 0

There are several ways of solving this equation. It is possible to solve for the position of this quadratic’s vertex, and demand that its y-coordinate is always greater than 0. It is also well known (by none more so than GCSE students) that a quadratic has no real roots if its discriminant is less than 0. The expression for the solutions of the equation

at^2+bt+c=0

is

\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

It is the quantity b^2-4ac that we call the discriminant, and we can see that if it is less than 0, no real solutions exist.

Then in our case we demand that

[-3gv\sin\theta]^2-4[g^2][2v^2]<0

yielding

\displaystyle \sin^2\theta<\frac{8}{9}

Hence

\displaystyle \theta < \arccos\bigg(\frac{1}{3}\bigg) \approx 70.5^\circ

Results like this make me glad I do physics.

If you throw a stone below this angle, it is always moving away from you. Throw a stone above this angle, and the stone will spend part of its journey moving towards you.

Throw a stone at this angle, and there will exist a single time t' at which the stone instantaneously moves along an imaginary circle with you at its centre and radius r(t').

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