This article in the Economist magazine sparked a physics debate in my pub, on the topic of straight-line trajectories: My protagonist was arguing that any body in a gravitational field will be subject to downward forces, no matter how fast it was going horizontally. His beermat calculation was (doubling the speed of the 30k shot) 4000kph = 1,111.1 m/s and therefore a travel time of 27 seconds over 30km. With a gravitational constant of 9.81 vertical falling distance ought to be 9.81*27*27*0.5 or 3,575m. Rather a lot. But alas, the earth is not vacuous. So I plugged some numbers in to this terminal velocity calculator here and a 10kg mass, 0.02 cross-section, 0.3 drag co-efficient gives me around 140m/s. Lets assume 5 seconds to reach terminal velocity and decent is additive after, 140*22 still gives me over 3km vertical decent. So, my question is, how can our beermat estimates (and assumptions) be so at odds with the statement in the article? This thread https://www.physicsforums.com/showthread.php?t=281635 says - accepted, and I'm not asking for anything exact, something else is missing here. Could Mr Finkenaur be referring to a non-Euclidean spherical geometry straight line that meant the slug was always 20ft (say) off the ground and in effect in orbit? I'd go with that kind of 'straight-line'. But orbital velocity is way over atmospheric speeds. Side note: A Distance to the Horizon Calculator tells me that I'd need to be at least 18m above the ground in order to fire 30km tangentially at the earth and the target would also have to be 18m above the ground. Doesn't suggest its useful against tanks if truly Euclidean.