If the launch angle is further known (for instance, by analyzing the frames of a video tape), then a more exact estimate can be obtained: $$E = {m g d \over 2 sin(2 \theta)}$$
The projectile distance equation for flat ground is: $$d = {v_0^2 \over g} sin(2 \theta)$$
The maximum projectile distance occurs with a launch angle of 45°, so we can omit the launch angle by using an inequality: $$d ≤ {v_0^2 \over g} sin(2 \cdot 45 ^{\circ} )$$ $$d ≤ {v_0^2 \over g}$$
Solve for \(v_0\): $$v_0 ≥ \sqrt {g d}$$
Plug that into the kinetic energy equation: $$E = {1 \over 2} m v_0^2$$ $$E ≥ {1 \over 2} m g d$$
An example query for Google Calculator is .5 * (9.8 m / s^2) * 100 grams * 100 meters.
| Slinger | Date | Sling type | Type | Mass | Throw style | Sling length | Range | Energy |
|---|---|---|---|---|---|---|---|---|
| Melvin Gaylor * | 1970 | 212.6g | 349.6m | ≥364J | ||||
| Vernon Morton * | 283.5g | 258.2m | ≥358J | |||||
| SEB | Stone | 300g | Side-Arm | 130cm | ~220m | ≥325J | ||
| Colonel Walker | Orange | ~454g | Modified Underhand | 122cm | ~130m | ≥290J | ||
| Douglas | 02/11/05 | Heavy stone | ~500g | ~90m | ≥220J |