This is where I work

So we have some message boards at work and I thought I would share one post that was on a Math board we have:

A fort, full of pirate-hating ninjas, has just received word that there is a pirate ship 50 km away. As luck would have it, the ninjas have a rail gun mounted at sea level that can fire a shot at mach 45 (14850 m/s). Because this problem isn’t absurd enough, the pirate ship has a hyperspace drive that is about to activate in 4 seconds and allow the pirates to escape. The pirate ship is not moving. Assuming that the world is round with a radius of 6300km and ninjas can change the gravitational force of the earth when fighting pirates, what level of gravity is necessary to hit the pirates at the water line of the ship if the ninjas fire the rail gun parallel with the ground.

Pirate ship
Ninja fort
50 km along surface of the sea
Not to scale
Ninja shot trajectory without gravity

and of course a solution was quickly posted

Solution:
I used 3.14 as an approximation for pi in the following calculations

We can consider the fort and ship to be on a single circle of radius 6300km, or a circumference of 39564km.
50/39564 * 360 = 0.45496 degrees for the angle of the arc between the two.

Since the Ninja’s are firing along the tangent to the circle, we have a right triangle with one leg of 6300km, adjacent to an angle of .45496.
6300/(cos (0.45496)) = 6300.1986 for the hypotenuse of the triangle, so with zero gravity, the projectile would be 198.6 meters above the pirate ship. (6300.1986 – 6300 = 0.1986 km)

At mach 45, it takes 3.367 seconds for the projectile to travel 50 km (50000/14850 = 3.367 seconds)

Then we can use the following formula to get the acceleration from gravity necessary for the projectile to strike the pirate ship.
½*at2 + v0t + x0 = 0, where a is the acceleration we are looking for, t is 3.367seconds, vo = 0, and x0 = 198.6 meters
Solving for a we get:
5.6683445a seconds2 = -198.6 m
a = -35.0366 m/s2

Edit: I just noticed I didn't compensate for the fact that while travelling the 50km arc, the projectile actually moves less than 50km in the tangental vector, and its final velocity at impact is actually faster than 14850 m/s, so the total time would be less than 3.367 seconds. However, the perpendicular vectors of the velocity and the slightly shorter distance are so small, that they shouldn't have any major effect on the final result. In any case, if my gravity number is just slightly high, the projectile going at mach 45 would probably just skip across the surface of the water and hit the boat anyway (and the shockwave of it hitting the water would cause some major waves.)

Awesome.