what i mean is that when say, an artillery shell is fired, or a piece of debris from an explosion is launched, it goes up and away at an almost 45 degree angle until its upwards momentum is no longer sufficient to fight gravity before dropping pretty much straight down. second, given the angle of impact on the ground, could you calculate backwards from there the initial velocity and launch site of the projectile?
In: 0
It’s an arc because gravity is acceleration, 9.8 meters per second squared. You drop something, and after one second it’s going 9.8 meters per second. After two seconds, it’s going 19.6 meters per second.
Shove a ball off a high place in a vacuum, and the forward velocity will be constant while the downward velocity keeps increasing, which creates the arc. And in real life we have air resistance, so that forward velocity will be reduced while the downward velocity increases, producing even more of an arc. Until we hit terminal velocity of course.
>second, given the angle of impact on the ground, could you calculate backwards from there the initial velocity and launch site of the projectile?
Yes, the military uses this for what’s called counter-battery fire. With the current technology and radar tracking of incoming rounds, we can have a return artillery shell in the air before the incoming shell lands. Thus it’s dangerous for anyone shooting at us to remain in one place, so “shoot and scoot.”
Latest Answers