[balloon-makers] Corrected Typo - Terminal Descent

Tim Baggett tim at oasis.com
Mon Mar 11 11:03:41 CST 2002



-----Original Message-----
From: Mike & Tammie <skyfun at ncia.net>
>How thick does the air have to be for it to take 6.25 seconds to reach
a
>velocity of 6.25 meters per second. The distance 1.99 meters is correct
but
>taking 6.25 seconds to get there....... the air has just turned into
>molasses.
>Had to be a mistype.
>> t = v / a = 6.25 m/s / 9.81 m/s^2 = 6.25 seconds


You are absolutly correct Mike, and it is a typo. I simply copied down
the same number from the velocity as the time to fall. The equation is
actually correct though. 9.81 / 6.25 comes out to .637 seconds, not 6.25
seconds.

Here it is with corrected typos:


1231 ft/sec = 6.25 meters/sec

Physics derivation
Known: d = 1/2 a * t^2
  d = distance moved
  a = acceleration
  t = time accelerated
first derivitive: v = a * t


Assume that when Tom jumps, there is no atmosphere. If there was, then
the air would cause friction and slow him down, making it take longer
for him to reach Keith's terminal descent velocity, or if the friction
is high enough he may never reach this target velocity. Good thing we
have an atmosphere however, since we ballooning would be impossible
without one.

Time needed to fall to reach 6.25 m/s (recall that acceleration of
gravity g=9.81 m/s^2 )

t = v / a = 6.25 m/s / 9.81 m/s^2 = 0.638 seconds

Tom needs to fall 0.638 seconds. How tall must the bridge be?

d = 1/2 g * t^2 = 1/2 * 9.81 m/s^2 * (0.638 s)^2 = 1.99 meters

1.99 meters comes to 6.54 feet.


Tim


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