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
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
Latest Answers