Using a sling whirled around above the head, could someone on the moon throw a stone into orbit?
tl;dr: probably deeply suborbital; 1192 meters, 38 seconds, either straight up or downrange at 45 degrees.
And of course even if you had superhuman sling skills, your trajectory would either fly off into space (lunar C3 > 0) or return on an elliptical (C3 <0) trajectory and intersect the surface somewhere. Standing on the Moon a horizontal throw at 1680 m/s would theoretically be in circular orbit at an altitude of 2 meters, but it would just slam into some crater edge or mountain.
@OrganicMarble's
The same speed as it could be thrown anywhere else? There's no $g$ in $\omega \times r$
is insightful but @SF.'s
The main limiting factor on Earth is air resistance. A stronger thrower could lob a heavier rock at similar speed, but the range of speeds is very limited by $v^2$ in air drag equation. On the Moon you could pump power into spin of the sling as long as you can control it and it doesn't break.
is important as well, and the "helicopter method" shown below does just that!
Scientific American's Whistling Sling Bullets Were Roman Troops' Secret Weapon says:
In the hands of an expert, a heavy sling bullet or stone could reach speeds of up to 100 mph (160 km/h): "The biggest sling stones are very powerful — they could literally take off the top of your head," Reid said.
That's 44 m/sec, and the Tod's Workshop video Is a sling as powerful as a gun? demonstrates a 125 ft/sec ~ 38 m/s measurement without trying very hard.
$\sqrt{GM_M/r_M}$ is about 1680 m/s so were only a few percent of orbital velocity.
Okay, but how suborbital would 44 m/s get you?
If we already know we're profoundly suborbital in terms of speed, we can estimate the maximum altitude (shooting straight up, probably killing yourself when it falls back on you) from conservation of energy, and maximum distance from a parabolic trajectory at 45 degrees.
Max height:
$$mv^2 = mgh$$
$$h = v^2/g$$
for 44 m/s and $g=GM_M/r_M^2=$ 1.62 m/s^2 gives 1191 meters!
The likelihood that it hits you on the way down is therefore small.
Wikipedia's Kinematic quantities of projectile motion give us
$$x(t) = v_0 t \cos\theta$$
$$y(t) = v_0 t \sin\theta - \frac{1}{2}g t^2$$
and solving for the time $t$ of the 2nd zero of $y$ and putting that back to solve for $x$ we get
$$x_2 = \frac{2 v_0^2}{g} \cos \theta \sin \theta$$
and using
$$\cos \theta \sin \theta = \frac{1}{2} \sin 2 \theta$$
$$x_2 = \frac{v_0^2}{g}\sin 2\theta$$
we see the familiar result that the maximum range is at $\theta = $ 45° and it will be again! a distance of $v_0^2/g =$ 1191 meters!
In both cases the flight time is about 38 seconds.
This will be much further than Allan Shephard's golf ball shot 1, 2
But seriously... in a Space Suit?
The truly stellar and fascinating video Slinging Target Practice - Six Techniques shows six slingshot techniques; Helicopter, Figure 8, Byzantine, Overhand, Underhand and Greek.
The "helicopter" technique seems best suited for suited astronauts. As long as the sling is not too long, you get it moving overhead with a smooth arm motion then just use small deflections to speed it up overhead.
You will be needing either some good wrist articulation from your suit, or have to make do with using elbow or shoulder to pump speed into the helicopter.
I think this could work, and will get you close to a 1 kilometer impact distance and 38 second flight time of your very suborbital projectile.
above: Helicopter technique, below: all six for comparison
