1. We generally model the high speed case of fluid resistance (typical
example: an object moving in air) by the quadratic approximation: f =
c v^2, where the constant (here: c) depends on the geometry of the
object and the density of the fluid.
2. We can't simply assume that the y-component of the velocity on impact
is the terminal velocity, and the x-component is the same as the
original horizontal velocity. Friction acts in the direction opposite
to the instantaneous velocity, so this is a vector differential
equation. With the common choice of "up" being the +y direction, and
choosing the +x direction to be perpendicular to it such that the
object's initial velocity has a component in the +x direction, we get:
-mg j_hat - c v^2 v_hat = m a_vector,
where v_hat is a unit vector in the direction of the velocity: v_hat =
v_vector/v, and j_hat is a unit vector in the y-direction (i_hat in the
x-direction).
-mg j_hat - c v^2 (v_x i_hat + v_y j_hat)/v = m (a_x i_hat + a_y j_hat),
or
-mg j_hat - c v (v_x i_hat + v_y j_hat) = m (a_x i_hat + a_y j_hat).
Separating the x- & y- components:
- c v v_x = m a_x and -mg - c v v_y = m a_y
If that second equation had (v_y)^2 instead of v v_y, we'd have our
normal 1-dimensional case (at least as far as the y-direction goes) with
v_y approaching sqrt(m/(cg)) asymptotically. But we don't. The length
of v_vector, v = sqrt(v_x^2+v_y^2), so the equations really are
- c v_x sqrt(v_x^2+v_y^2) = m a_x
and -mg - c v_y sqrt(v_x^2+v_y^2) = m a_y
The bad news is that both of these involve both x and y derivatives,
there's no way to separate variables and integrate. Numerical
approximations can be done quite easily -- for instance with an Excel
spreadsheet -- and the basic way I think of it is that we approach
terminal velocity in _some_ direction of motion, while the angle is also
approaching vertical downward.
Give it enough time and the object will be heading basically straight
downward, basically at the terminal velocity v_t = sqrt(m/(cg)).
(Notice that the formula for v_t depends on which approximation we're
using to model the fluid resistance! They're all approximations,
anyway.)
Fun to think about this stuff!
Ken Caviness
Physics
Southern Adventist University
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Roger
Haar
Sent: Wednesday, January 24, 2007 10:52 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Final velocity of bullets
Hi,
I am 99% sure that a few years ago in Phoenix, there was serious
or
fatal injury from a celebratory near vertical gun shot. As I recall,
the bullet entered the top of the skull. Since then there has been a
media campaign to warn people not to do this and the police have
attempted to clamp down. Thus it is known experimentally that
vertically fired bullets can be deadly.
**********************************************************************
From Michael D. Edmiston, Ph.D.
(a) At what elevation must a typical gun be fired such that air friction
reduces the velocity sufficiently that the landing velocity is near
free-fall terminal velocity? I think we are agreed that a bullet fired
straight up will land at terminal velocity. I think we are also agreed
that a bullet fired at steep angle, but not straight up, will land with
some horizontal velocity in addition to terminal vertical velocity. Is
there an angle of launch above which a typical bullet is not lethal.
(And I think we realize this is gun and bullet dependent.)
I am no so sure that one can decompose air friction into
components
like that. I think the horizontal velocity is going to effect the
vertical terminal velocity. The bullet/ air interaction is more
turbulent than viscous in nature so the drag force is proportional to
v^2 not v.