What coilgun exit speed is needed to launch a projectile 16' straight up?
Short Answer: The same speed as a projectile that falls 16' straight down.
Long Answer: Solve for velocity v when distance d=16'. Use the basic equations for velocity and distance as a function of time under a constant acceleration g:
d = gt2 / 2 and v = gt.
Solve the second equation for t = v/g and substitute into the first. Re-arrange and simplify to get the general equation for velocity v of a falling body as a function of height (distance). The result is:
v = sqrt( 2 x g x d )
Substitute gravity g = 32 ft/sec/sec and height d = 16 ft, the necessary launch speed v = 32 ft/sec which is the same as 9.8 m/s.
Last update May 7, 2007 by Barry Hansen ©1998-2007