In this situation the cannon is being shot towards the ground from a blimp (fantasy situation) .
Blimp's speed is 35 mph.
Blimp's cruising ceiling is 8000 ft.
Standard Civil War cannonball was 12 lbs (5.4 kg).
The shot speed is 250 miles per hour (820 ft per second).
The cannons wouldn't be pointed vertically obviously but straight out of the sides of the ship, probably a 45 degree angle.
How far away would the ship have to be to hit any spot, and how long would the shot take to reach the ground from 8,000 ft up? Would it be more realistic to have the guns pointed more towards the ground?
Thank you :)
Assuming no air resistance, the vertical component of the cannonball's initial velocity in meters per second is $250\sin(-\frac\pi4) \approx 176.8.$ (The angle $\frac\pi4$ is $45$ degrees converted to radians). In the absence of air resistance, when the ball hits the ground it would have a kinetic energy equal to its initial kinetic energy plus the work done by gravity as the ball falls $2400$ meters. Since the vertical motion of a ball fired at an angle is independent of the horizontal velocity (in the absence of air resistance), we can assume zero horizontal velocity for the moment, in which case the initial kinetic energy for a cannonball of mass $m$ is $\frac12mv^2 = \frac12 m(250\sin(-\frac\pi4))^2 = 15625m$ and the work done by gravity is $mgh = m 9.8\cdot2400 = 23520m,$ for a final kinetic energy of $39145m,$ which gives a velocity of about $279.8.$ Average vertical velocity $\bar v$ (averaged over time) between the cannon and the ground is therefore approximately $\frac12(176.8 + 279.8) = 228.3,$ and the time would be $h/\bar v \approx 10.51,$ measured in seconds.
Putting the horizontal velocity back to the actual $250\cos(\frac\pi4),$ the horizontal distance traveled would be that velocity times the elapsed time until impact, which works out to about $1860$ meters.
But it seems unrealistic to ignore air resistance for a $12$-pound ball traveling at these kinds of speeds for so long. Putting $5.4$ kg and $0.01$ m$^2$ into http://www.calctool.org/CALC/eng/aerospace/terminal I get a terminal velocity of $147$ m/s. I don't know how quickly the cannonball would decelerate to this speed, but at the initial velocity of $250$ meters per second, using a coefficient of drag of $0.4$ I get a drag force of about $150$ newtons, which for a $5.4$-kg cannonball is a deceleration of almost three times the acceleration of gravity. I have not worked this out in simulation, but the initial deceleration (if it continued) would bring the cannonball to a complete stop long before it hit the ground, so my initial guess would be that the cannonball would decelerate to terminal velocity (or near terminal velocity) before it hit.
If that guess is correct, you would do just about as well simply dropping the cannonball onto the target below rather than shooting it. Moreover, by leaving the cannon off the blimp you have that much more capacity to carry extra ammunition.