Do the classics still work? and horizons

[Phymath]

in a case were we use a lot of inital velocity, does the projectiles equations
($$x = x_0 + v_0t + \frac{1}{2}at^2$$) still hold I mean when its range is so large that it actually will go beyond the horizon? How does the equation compensate for a rounded surface or does it not have to? btw how far is the horizon for an average humans height, how do u figure that out for a given height?

 

[Clausius2]

I would say cartesian coordinates are not valid when the curvature of the Earth is taken into account into the projectile equations. Surely $$\overline{g}$$ will not point towards the "-y" direction but "-r" direction, and both x and y coordinates will have a component of $$\overline{g}$$

Anyway, you can model such projectile with polar coordinates, no matter what initial velocity has. In the limit in which the radius of curvature of the ground tends to zero (small trajectories), the polar coordinates collapses over cartesian coordinates.

 

[T@P]

The projectile would not follow a strictly parabolic curve if you go *very* far out since the acceleration would not be in the same direction. An extreme example would be an orbiting asteroid, but in this case it would follow a circle-ish shape. I think those problems would be a lot easier to solve using work and energy, just a thought.

 

[marcus]

btw how far is the horizon for an average humans height, how do u figure that out for a given height?

Phymath that is a nice question and IIRC the answer is like this.

To a good approx, on a sphere, when you are height h above surface,
the distance X from your eye to horizon is the solution to this:

X/h = D/X, where D is diameter of sphere

So let us say diameter of Earth is 8000 miles and you are 125 miles above surface, so

X/125 = 8000/X

then to solve
X² = 1000 000
X = 1000 miles

that is, if you are 125 miles high then you can see 1000 miles

or, if you are 1.25 miles high, then you can see 100 miles.

Another way to write formula is

X² = Dh = 2Rh
where h is height of observer and R is radius of sphere.

the proof is by pythagoras triangle formula and simple picture
and an approx which depends on h being small compared with R.

 

[Phymath]

thanks man!