The simplest would be defining some arbitrary impact velocity that is at the limit of being fatal, and we then consider everything else (surface properties, subject's physique,...) except gravitational acceleration constant. We can also neglect air resistance to make it simpler, since we're more interested in a safe height to jump off on the Moon, than that on the Earth. Plus air resistance at such small heights wouldn't change our results much anyway.
Let's, for the sake of argument, assume we have a crash test dummy that won't break into pieces, if it impacts the ground no faster than 60 km/h. At Earth's mean sea-level surface gravitational acceleration (9.80665 m/s2) that comes out as a jump at a height of 14.16 m, and the free fall duration of 1.7 s before impact. I'm cheating by using an online free fall and constant acceleration calculator, but the maths for a free fall without air resistance go as follows:
$$v(t)=-gt+v_{0}$$ $$y(t)=-\frac{1}{2}gt^2+v_{0}t+y_0$$
where
- $v_{0}$ is the initial velocity (m/s).
- $v(t)$ is the vertical velocity with respect to time (m/s).
- $y_0$ is the initial altitude (m).
- $y(t)$ is the altitude with respect to time (m).
- $t$ is time elapsed (s).
- $g$ is the acceleration due to gravity (9.81 m/s2 near the surface of the earth).
We can derive everything else we need from these two equations, let me know if you require a more detailed answer in this part. Moving on, now let's see at which height of a jump we reach 60 km/h impact on the surface of the Moon. Moon's average surface gravity acceleration is 1.622 m/s2. Plugging that into our online constant acceleration calculator (inserting acceleration, start velocity of 0 km/h and end velocity of 60 km/h), we get:
- Jump from a height of 85.63 m
- Free fall time of 10.28 s
Recalculation for 110 mph (177 km/h) impact velocity post edit of the question:
- Jump from a height of 745.41 m (2,445.58 ft) on the Moon, 123.29 m (404.49 ft) on Earth
- Free fall time of 30.32 seconds on the Moon, 5.01 seconds on Earth
And for 116 mph (187 km/h), equaling 4,000 N force on impact of a 170 lbs (77.11 kg) object:
- Jump from a height of 829.49 m (2,721.42 ft) on the Moon, 137.2 m (450.12 ft) on Earth
- Free fall time of 31.98 seconds on the Moon, 5.29 seconds on Earth
Edit to add: Both these last impact velocities (110 and 116 mph) are over 80.85 mph (130.12 km/h) that I calculate (see comments for details, thanks @LorenPechtel!) is the terminal velocity on the Moon for a freeflying skydiver, if all other conditions remain constant (mass of the skydiver, drag coefficient, fluid density, projected area of the falling object) both on the Earth and on the Moon. I.e. this would mean equal atmospheric conditions, no spacesuit, and for a skydiver whose terminal velocity on Earth is 320 km/h (200 mph or 90 m/s), as given in Wikipedia examples of terminal velocity.
So if we accept that a human can survive an impact at given velocities (110 and 116 mph), Moon's gravity isn't sufficient to counteract atmospheric drag of 1 atmospheric pressure to such speeds to kill you on impact. I.e. the maximum survivable altitude to jump off in such conditions would be, theoretically, infinite. In reality, you'd then have to deal with the heat released during atmospheric reentry beyond terminal velocity, which becomes a different question altogether. And it would of course be unfeasible, the Moon doesn't have sufficient gravity and magnetic field to sustain Earth-like atmospheric pressure.
Solution to calculate lunar terminal velocity, with a known terminal velocity of the same object and same atmospheric density on Earth:
$$v_{(M)} = 0.4067 \cdot v_{(E)}$$
Where $v_{(E)}$ is terminal velocity for Earth's gravity. Derived from:
$$v= \sqrt{\frac{2mg}{\rho A C_d }}, \ x = \frac{2m}{\rho A C_d }$$
Where $x$ denotes our constant mass of the object, drag coefficient, fluid density, and projected area of the falling object. Thus:
$${v_{(M)}}^2 = \frac{g_{(M)}{v_{(E)}}^2}{g_{(E)}} \Rightarrow {v_{(M)}}^2 = \frac{1.622}{9.80665} \cdot {v_{(E)}}^2 \Rightarrow v_{(M)} = \sqrt{0.1654 \cdot {v_{(E)}}^2}$$
For a belly-to-earth (i.e., face down) free-fall position, again using Wikipedia quoted values, in Earth's gravity achievable terminal velocity is only 195 km/h (122 mph or 54 m/s). On the Moon, using our conversion value of 0.4067, this amounts to 79.3 km/h (49.6 mph or 22 m/s). I would still argue that this is not survivable due to lack of vegetation and unweathered (sharp) terrain on the Moon, but it is a food for thought. For a 170 lbs, this "only" amounts to an impact force of 1,710 Newtons. With some luck and landing into a deep pocket of lunar dust and no larger boulders in the way, it might be survivable. Still, that's in a face down position, so most likely not.
Another thought though, that our subjects are acrobats that would probably achieve jumps of such heights with their own power. So they would first have to work against the same gravitational acceleration that will later try to kill them on impact. Point being, that if you can't jump on Earth to heights that would be fatal to land from, you won't be able to do that on the Moon either. So the frequency of injuries and fatalities should be much the same, assuming all other conditions are equal, and no severe bone density loss due to staying in roughly 1/6-th the Earth's gravity. Of course, bone density loss usually comes with muscle atrophy, so the ability to jump as high might decrease, too.