KICKMAN

That was a lot of drawn panels just to catch a guy. But just goes to show that Aquarium is not useless.

It stands to reason that a hero who is more powerful in water (and who dies if he’s out of water for too long) might know the locations of all the major pools, ponds, estuaries, reflecting pools, and commemorative fountains in Hero City. For emergencies.

I know we have some mathematicians and physics-ians that read Kickman, so I’ll let them double check my work here. Aquarium is my Aquaman analog–and according to Green Ronin’s DC Hero’s Handbook (which is canon because they wrote it in conjunction with DC Comics), Aquaman has a swimming speed of 500 mph. I asked Grok, “How far would a human body travel if propelled at 500 mph assuming it is launched at 20 degree angle to the ground?” It gave me a very long answer with several equations and came up with a little over 50 miles–but that assumes no air resistance. So I’d say using a few pools to intercept a flying teammate is within the realm of reality. Well, comic book reality. In fact, if he had a direct path, he could have probably done it in one giant dolphin leap.

I stand by the page. Although, this was a lot of panels and a lot of drawing. Next page is just going to be a full-page close up of Sparta’s boobs.

To determine how far a human body would travel if propelled at 500 mph at a 20° angle to the ground, we model this as a projectile motion problem under ideal conditions: no air resistance (a common physics approximation for such calculations, especially since the query doesn’t specify drag), constant gravity (\(g = 32.2\) ft/s² or \(9.8\) m/s²), and launch from ground level with the goal of finding the horizontal range to where it lands back at the same height.

### Step 1: Convert initial speed to consistent units
– Initial speed \(v_0 = 500\) mph.
– Convert to feet per second:
\(v_0 = 500 \times \frac{5280 \text{ ft/mi}}{3600 \text{ s/h}} = 500 \times \frac{22}{3} \approx 733.333\) ft/s.
(Exact: \(v_0 = \frac{500 \times 22}{3} = \frac{11000}{3}\) ft/s.)

### Step 2: Resolve initial velocity into components
– Horizontal: \(v_{0x} = v_0 \cos \theta = v_0 \cos 20^\circ\).
– Vertical: \(v_{0y} = v_0 \sin \theta = v_0 \sin 20^\circ\).

### Step 3: Derive the range formula
The time of flight \(t\) is the duration until the projectile returns to the initial height (symmetric trajectory):
\(t = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin \theta}{g}\).

Horizontal range \(R =\) distance traveled horizontally:
\(R = v_{0x} \times t = (v_0 \cos \theta) \times \frac{2 v_0 \sin \theta}{g} = \frac{v_0^2 \sin 2\theta}{g}\).
(This uses the double-angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\); maximum range occurs at \(\theta = 45^\circ\).)

### Step 4: Compute numerically
– \(\sin 40^\circ \approx 0.6428\) (since \(2\theta = 40^\circ\)).
– \(g = 32.2\) ft/s².
– \(v_0^2 = \left( \frac{11000}{3} \right)^2 = \frac{121000000}{9}\).
– Numerator: \(v_0^2 \sin 40^\circ = \frac{121000000}{9} \times 0.6428 \approx 8639135.555\).
– \(R = \frac{8639135.555}{32.2} \approx 268290\) feet.
– Convert to miles: \(268290 \div 5280 \approx 50.81\) miles.

(Exact computation yields approximately 50.81 miles; with more precise \(\sin 40^\circ = \sin(40 \times \pi/180)\), it’s consistent.)

The body would travel approximately **50.8 miles** (or about 81.8 km) horizontally before landing.

In reality, air resistance would drastically reduce this distance at such high speeds (terminal velocity for a human is ~120–200 mph, and drag forces would dominate), but the ideal projectile calculation gives this result.