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 *can* go away at 45°, but it doesn’t have to. A projectile can follow a parabolic arc with an initial angle anywhere between 0° and 90°.
Edit: technically it could be fired from an elevated position with a negative angle, and still follow a parabolic arc.
Yes, that’s possible. It’s exactly what ballistic experts from forensic crime scene units do sometimes.
Yes, you can and there are anti-artillery weapons systems based on this concept:
[https://en.wikipedia.org/wiki/Counter-battery_radar](https://en.wikipedia.org/wiki/Counter-battery_radar)
As another comment noted, ballistic trajectories aren’t limited to starting at 45 degrees.
Yes. Projectile motion is one of the early things someone studying physics typically looks at.
Ignoring things like air resistance and the curvature of the earth, the main force acting on the projectile after projection is gravity. If you ignore gravity for a moment, the projectile would just continue in a straight path forever (again, ignoring things like air resistance). With gravity, there is a gravitation force that accelerates the projectile back towards the ground. If you break the motion of the projectile into two axes, one parallel and one perpendicular to the ground, the projectile will continue to travel in the direction parallel to the ground at the projection velocity (again, ignoring things like air resistance) until the projectile collides with the ground from gravity accelerating it in that direction. The trajectory that this ends up being is a parabola.
a_y(t) = -g (y acceleration = acceleration due to gravity, constant)
Integration gives:
v_y(t) = v_y(0) – g*t (y velocity at time t = initial y velocity + acceleration due to gravity * time)
Integration gives:
y(t) = y(0) + v_y(0)*t – 1/2*g*t^2 (y position at time t = initial y position + initial y velocity * time + 1/2 * (acceleration due to gravity)^2)
x is simpler:
a_x(t) = 0 (no acceleration in x axis, ignoring air resistance)
v_x(t) = v_x(0)
x(t) = x(0) + v_x(0)*t
If you have either the initial conditions or the end conditions, you could solve this system of equations and plot it which would yield a parabola!
For instance, assume initial conditions of:
– x(0)=0m
– y(0)=0m
– v_x(0)=5m/s
– v_y(0)=5m/s
– g = 9.8m/s^2
This is the projectile starting at the origin (0,0) with an initial velocity of 5m/s in both the x and y directions (45° initial trajectory):
Solution and graph: https://www.wolframalpha.com/input?i=x%3D0%2B5*t%2C+y%3D0%2B5*t-1%2F2*9.8*t%5E2 (copy/paste the whole URL)
See https://en.wikipedia.org/wiki/Projectile_motion
You answered your question completely.
Gravity pulls on all objects the same.
Step 1: load the shell. The shell is in the gun and the bolt is locked. The shell isn’t moving. The shell IS accelerating toward the center of the earth at 9.8 meters per second, every second (9.8 m/s/s, or 9.8 m/s²).
Step 2: bang! The shell is accelerated by the charges like 1 million meters per second for a tiny fraction of a second. It reaches it’s highest speed right at the end of the barrel.
The moment it leaves the barrel it starts slowing down because gravity is pulling it down at 9.8 m/s².
Eventually, gravity steals enough energy that the bullet can’t continue going up, gravity takes over, and the shell comes down.
The math you’re talking about is also used during aiming. They know the distance to the target. They can set how hard they shoot the shell, but only in increments.
Have you seen artillery loading? The shell goes in first then they load bags of explosives. Each bag equals the same projectile force and you can load up to 5. Loading only 2 means the shell can’t go as far as loading 5.
So they know how far they need to shoot, how heavy the shell is, and how far each bag will get them. The only thing left to determine is the launch angle.
They have such a distinctive arc because air friction inertia and gravity affects all bullets equally-ish. All free fall trajectories also have a distinctive straight line trajectory for the exact same motive.
And yes, based on the size/weight of the bullet, angle of impact and damage on the ground you can calculate initial velocity, angle and distance.
Any unpowered object flying through the air will travel with the combination of 2 formulas: the vertical axis, which is made up of how much velocity it has vertically subtracted by the acceleration of gravity; and the horizontal axis, which is made up of how much velocity it has horizontally. If we ignore the real world things (like air resistance) this will always create a parabola.
So because parabolas are very well studied shapes (especially with calculus), if we know the shape of one part of it and the acceleration it has been going through (aka gravity, so yeah we know it) we can backtrack the shape of it at any other part.
What you looking for is Classical Physics, specifically Projectile Motion.
In general, you have a projectile with some initial velocity, and through out its travel, gravity (and sometime friction if it isn’t negligible) apply a force on it. Knowing the initial conditions let you calculate how it will ended up, and vice versa.
In this case, if you know the angle that it hit the ground and how fast it going roughly based on the impact, you can approximate pretty close where it was fire initially.
If you want a deeper look, just google “Cannonball physics problem”
Because the formula for position of an object with acceleration and velocity has an x^2 term in it. Squares form parabolas when graphed.
>second, given the angle of impact on the ground, could you calculate backwards from there the initial velocity and launch site of the projectile?
With just the angle, no. You’d need some more information.
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