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Roger is correct on both counts:
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).
Writing a_vector = a_x i_hat + a_y j_hat, v_vector = v_x i_hat +
v_y j_hat, this gives
-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