Chronology | Current Month | Current Thread | Current Date |
[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |
velocity, it is always -g, irrespective of the launch angle.
In other
words, for a vertical trajectory the rate of schange of speed is
always -g, but once the trajectory has any angle at all different
from 90 degrees, the rate of change of speed is 0 instantaneously at
the top of the trajectory.
That's what the math says, unless I have set the problem up wrong.
But that doesn't square with the physics as I know it. Why should the
presence of a rate of change of speed at the top of the trajectory be
dependent on whether the trajectory was vertical or not? This says
that different inertial observers would see different rate of change
of speed at the top depending on whether or not their speed relative
to the projectile was v(0)cos(theta) or not. That makes no sense to
me at all.
This says
that different inertial observers would see different rate of change
of speed at the top depending on whether or not their speed relative
to the projectile was v(0)cos(theta) or not. That makes no sense to
me at all.