Who can answer this? (gravity) [Archive]

David Williams

Apr20-04, 01:34 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>-> Hi everyone,\n-> This is my first post.\n-> I was talking with a few people about the chance of survival when a\n-> person falls from a 230 foot bridge into water. I\'m in the Coast Guard\n-> in NY Habor and we deal with people jumping from bridges, and a few of\n-> us wanted to know how fast a person would be hitting the water. I\n-> would be happy to do the math I just need the formula.\n\nIs the person wearing a parachute?\n\nIf there\'s no air resistance (which a parachute would magnify), the\nformula you want is:\n\nv^2 = 2 x g x h\n\nwhere v is the speed at which the body hits the water, g is the\nacceleration due to gravity, and h is the height of the bridge. In the\nkinds of units you appear to like, g is 32 ft/second^2, h would be in\nfeet, and v in feet per second.\n\nAny real person would be somewhat slowed down by air resistance, but\nnot much, without a parachute.\n\ndow\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>-> Hi everyone,
-> This is my first post.
-> I was talking with a few people about the chance of survival when a
-> person falls from a 230 foot bridge into water. I'm in the Coast Guard
-> in NY Habor and we deal with people jumping from bridges, and a few of
-> us wanted to know how fast a person would be hitting the water. I
-> would be happy to do the math I just need the formula.

Is the person wearing a parachute?

If there's no air resistance (which a parachute would magnify), the
formula you want is:

v^2 = 2 x g x h

where v is the speed at which the body hits the water, g is the
acceleration due to gravity, and h is the height of the bridge. In the
kinds of units you appear to like, g is 32 ft/second^2, h would be in
feet, and v in feet per second.

Any real person would be somewhat slowed down by air resistance, but
not much, without a parachute.

dow

Doug Sweetser

Apr21-04, 03:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello:\n\nHow does one _find_ a formula? I met Paul Hewitt, the author of\n"Conceptual Physics." He recommends to always write down what you want\nin the end. In this case:\n\nv =\n\nPaul gives 40% credit for taking that critical first step in the\ndirection you want to go. The formula must depend on two things: the\nacceleration of gravity g, with units of m/s^2, and the height of the\nbridge, h, in m. If gravity were stronger, v would be greater. If the\nbridge were higher, v would be greater. So as a first guess at a\nformula, try:\n\nv =?= g h\n\nCheck the units, and, oops, g h is m^2/s^2. That is the units of\nvelocity squared, so a correct equation on dimensional grounds is:\n\nv =?= (g h)^(1/2)\n\nIf you stop here, the formula is 40% off. Not so bad :-) The one last\nbit to look for is a numerical constant that reflects something about\nthe geometry of what is going on. That 2 requires more explaining than\nthe g h :-) Really this is a balance of two forms of energy:\n\nkinetic: 1/2 m v^2 vs. gravity: m g h\n\nThe gravitational potential gets converted to kinetic energy. The\nformula assuming no wind resistance is:\n\nv = (2 g h )^(1/2)\n\nThis is not true in practice. Sky divers without parachutes slam into\nthe ground at 120 MPH regardless of how high the initial jump was. You\ncan ask: will this person reach terminal velocity from that bridge?\n\n\ndoug\nquaternions.com\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello:

How does one _find_ a formula? I met Paul Hewitt, the author of
"Conceptual Physics." He recommends to always write down what you want
in the end. In this case:

v =

Paul gives 40% credit for taking that critical first step in the
direction you want to go. The formula must depend on two things: the
acceleration of gravity g, with units of m/s^2, and the height of the
bridge, h, in m. If gravity were stronger, v would be greater. If the
bridge were higher, v would be greater. So as a first guess at a
formula, try:

v =?= g h

Check the units, and, oops, g h is m^2/s^2. That is the units of
velocity squared, so a correct equation on dimensional grounds is:

v =?= (g h)^(1/2)

If you stop here, the formula is 40% off. Not so bad :-) The one last
bit to look for is a numerical constant that reflects something about
the geometry of what is going on. That 2 requires more explaining than
the g h :-) Really this is a balance of two forms of energy:

kinetic: 1/2 m v^2 vs. gravity: m g h

The gravitational potential gets converted to kinetic energy. The
formula assuming no wind resistance is:

v = (2 g h )^(1/2)

This is not true in practice. Sky divers without parachutes slam into
the ground at 120 MPH regardless of how high the initial jump was. You
can ask: will this person reach terminal velocity from that bridge?

doug
quaternions.com

Danny Ross Lunsford

Apr21-04, 03:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>David Williams wrote:\n\n> If there\'s no air resistance (which a parachute would magnify), the\n> formula you want is:\n>\n> v^2 = 2 x g x h\n>\n> where v is the speed at which the body hits the water, g is the\n> acceleration due to gravity, and h is the height of the bridge. In the\n> kinds of units you appear to like, g is 32 ft/second^2, h would be in\n> feet, and v in feet per second.\n>\n> Any real person would be somewhat slowed down by air resistance, but\n> not much, without a parachute.\n\nWell air resistance is a huge factor. The terminal velocity for a\nfalling person is somewhere in the range 120-150 mph. A practiced\nskydiver can streamline his profile and achieve up to 180 mph.\n\n-drl\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>David Williams wrote:

> If there's no air resistance (which a parachute would magnify), the
> formula you want is:
>
> v^2 = 2 x g x h
>
> where v is the speed at which the body hits the water, g is the
> acceleration due to gravity, and h is the height of the bridge. In the
> kinds of units you appear to like, g is 32 ft/second^2, h would be in
> feet, and v in feet per second.
>
> Any real person would be somewhat slowed down by air resistance, but
> not much, without a parachute.

Well air resistance is a huge factor. The terminal velocity for a
falling person is somewhere in the range 120-150 mph. A practiced
skydiver can streamline his profile and achieve up to 180 mph.

-drl

Dirk Bruere at Neopax

Apr21-04, 03:24 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>nhilarzcyk wrote:\n\n> Hi everyone,\n> This is my first post.\n> I was talking with a few people about the chance of survival when a\n> person falls from a 230 foot bridge into water. I\'m in the Coast Guard\n> in NY Habor and we deal with people jumping from bridges, and a few of\n> us wanted to know how fast a person would be hitting the water. I\n> would be happy to do the math I just need the formula.\n\nSomeone else has provided the formula in abscence of air resistance.\n\nHowever, survival is more complex.\nIf someone hits flat, they are dead.\nIf they go in perfectly vertical head or feet first, nice and tight, they\n*might* live. IIRC the high diving record (non fatal and without serious\ninjury)is 170ft.\n\nHowever, if they adopt the correct streamlined posture they are going to hit\nfaster than if they were falling in a flat position.\n\nTheoretically for best \'performance\' they need to fall flat, and then rotate to\nenter the water vertically. A neat trick if you can do it.\n\nThen we have the question of clothing acting as a brake - does it protect or\ninjure since it slows the body faster in both air and water esp if it balloons?\nAnyone here dived both with and without clothing?\n\nBasically, it is a very complex question aerodynamically and I suspect that\nthere exists no simple equation which will give a real-life answer.\n\nSo, how many people *have* survived that 230ft fall?\n\n--\nDirk\n\nThe Consensus:-\nThe political party for the new millenium\nhttp://www.theconsensus.org\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>nhilarzcyk wrote:

> Hi everyone,
> This is my first post.
> I was talking with a few people about the chance of survival when a
> person falls from a 230 foot bridge into water. I'm in the Coast Guard
> in NY Habor and we deal with people jumping from bridges, and a few of
> us wanted to know how fast a person would be hitting the water. I
> would be happy to do the math I just need the formula.

Someone else has provided the formula in abscence of air resistance.

However, survival is more complex.
If someone hits flat, they are dead.
If they go in perfectly vertical head or feet first, nice and tight, they
*might* live. IIRC the high diving record (non fatal and without serious
injury)is 170ft.

However, if they adopt the correct streamlined posture they are going to hit
faster than if they were falling in a flat position.

Theoretically for best 'performance' they need to fall flat, and then rotate to
enter the water vertically. A neat trick if you can do it.

Then we have the question of clothing acting as a brake - does it protect or
injure since it slows the body faster in both air and water esp if it balloons?
Anyone here dived both with and without clothing?

Basically, it is a very complex question aerodynamically and I suspect that
there exists no simple equation which will give a real-life answer.

So, how many people *have* survived that 230ft fall?

--
Dirk

The Consensus:-
The political party for the new millenium
http://www.theconsensus.org

Paul Danaher

Apr21-04, 03:25 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>nhilarzcyk wrote:\n> Hi everyone,\n> This is my first post.\n> I was talking with a few people about the chance of survival when a\n> person falls from a 230 foot bridge into water. I\'m in the Coast Guard\n> in NY Habor and we deal with people jumping from bridges, and a few of\n> us wanted to know how fast a person would be hitting the water. I\n> would be happy to do the math I just need the formula.\n\nThe key google term here is "terminal velocity" - to quote\nhttp://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l3e.html\n\nAs an object falls, it picks up speed. The increase in speed leads to an\nincrease in the amount of air resistance. Eventually, the force of air\nresistance becomes large enough to balances the force of gravity. At this\ninstant in time, the net force is 0 Newtons; the object will stop\naccelerating. The object is said to have "reached a terminal velocity." The\nchange in velocity terminates as a result of the balance of forces; the\nvelocity at which this happens is called the "terminal velocity."\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>nhilarzcyk wrote:
> Hi everyone,
> This is my first post.
> I was talking with a few people about the chance of survival when a
> person falls from a 230 foot bridge into water. I'm in the Coast Guard
> in NY Habor and we deal with people jumping from bridges, and a few of
> us wanted to know how fast a person would be hitting the water. I
> would be happy to do the math I just need the formula.

The key google term here is "terminal velocity" - to quote
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l3e.html

As an object falls, it picks up speed. The increase in speed leads to an
increase in the amount of air resistance. Eventually, the force of air
resistance becomes large enough to balances the force of gravity. At this
instant in time, the net force is Newtons; the object will stop
accelerating. The object is said to have "reached a terminal velocity." The
change in velocity terminates as a result of the balance of forces; the
velocity at which this happens is called the "terminal velocity."

John C. Polasek

Apr22-04, 02:54 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Tue, 20 Apr 2004 06:34:38 +0000 (UTC), david.williams@bayman.org\n(David Williams) wrote:\n\n>-> Hi everyone,\n>-> This is my first post.\n>-> I was talking with a few people about the chance of survival when a\n>-> person falls from a 230 foot bridge into water. I\'m in the Coast Guard\n>-> in NY Habor and we deal with people jumping from bridges, and a few of\n>-> us wanted to know how fast a person would be hitting the water. I\n>-> would be happy to do the math I just need the formula.\n>\n>Is the person wearing a parachute?\n>\n>If there\'s no air resistance (which a parachute would magnify), the\n>formula you want is:\n>\n>v^2 = 2 x g x h\n>\n>where v is the speed at which the body hits the water, g is the\n>acceleration due to gravity, and h is the height of the bridge. In the\n>kinds of units you appear to like, g is 32 ft/second^2, h would be in\n>feet, and v in feet per second.\n>\n>Any real person would be somewhat slowed down by air resistance, but\n>not much, without a parachute.\n>\n> dow\nA handy formula is v = 8*sqr(h), so with a 225\' height, it\'s 8 x 15 =\n120 ft/sec.\n\nMr. Dual Space\n(If you have something to say, write an equation.\nIf you have nothing to say, write an essay).\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 20 Apr 2004 06:34:38 +0000 (UTC), david.williams@bayman.org
(David Williams) wrote:

>-> Hi everyone,
>-> This is my first post.
>-> I was talking with a few people about the chance of survival when a
>-> person falls from a 230 foot bridge into water. I'm in the Coast Guard
>-> in NY Habor and we deal with people jumping from bridges, and a few of
>-> us wanted to know how fast a person would be hitting the water. I
>-> would be happy to do the math I just need the formula.
>
>Is the person wearing a parachute?
>
>If there's no air resistance (which a parachute would magnify), the
>formula you want is:
>
>v^2 = 2 x g x h
>
>where v is the speed at which the body hits the water, g is the
>acceleration due to gravity, and h is the height of the bridge. In the
>kinds of units you appear to like, g is 32 ft/second^2, h would be in
>feet, and v in feet per second.
>
>Any real person would be somewhat slowed down by air resistance, but
>not much, without a parachute.
>
> dow
A handy formula is v = 8*sqr(h), so with a 225' height, it's 8 x 15 =
120 ft/sec.

Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).

David Williams

Apr23-04, 03:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>-> -> Well air resistance is a huge factor. The terminal velocity for a\n-> -> falling person is somewhere in the range 120-150 mph. A practiced\n-> -> skydiver can streamline his profile and achieve up to 180 mph.\n->\n-> -> -drl\n->\n-> I think he said the bridge was 230 feet high. Let\'s call that 225,\n-> which is 15^2. So, neglecting air resistance, the body would hit the\n-> water at sqrt(64 x 225) or 120 feet per second. That\'s about 83 miles\n-> per hour - considerably less than the terminal velocity for a person\n-> without a parachute. Since air resistance is roughly proportional to\n-> the square of velocity, even at the end of the fall the resistance\n-> would be only a small fraction of the weight. For most of the fall, it\n-> would be only a *very* small fraction.\n->\n-> In this case, air resistance would slow the fall only slightly.\n->\n-> dow\n\nFor the heck of it, I wrote a few lines of BASIC that simulate the\nfall, taking air resistance into account. I assumed a terminal velocity\nof 200 feet per second, which is about 135 mph. I\'ll append the code\nbelow. The result is that the body hits the water at 111 ft/sec, or 76\nmiles per hour.\n\ndow\n\n---------------------------------------------------\n\n\' bfall\ng = 32 \' ft/sec^2\nh = 230\' bridge height\nv = 0 \' speed of fall\ns = 0 \' distance fallen\nt = 200\' terminal velocity, ft/sec\ndt = .001\' iteration time interval, secs\nDO\nv = v + g * dt * (1 - (v / t) ^ 2)\ns = s + v * dt\nLOOP UNTIL s >= h\nPRINT v; "ft/sec, or";\nPRINT v * 3600 / 5280; "miles per hour"\nEND\n\n----------------------------------------------------\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>-> -> Well air resistance is a huge factor. The terminal velocity for a
-> -> falling person is somewhere in the range 120-150 mph. A practiced
-> -> skydiver can streamline his profile and achieve up to 180 mph.
->
-> -> -drl
->
-> I think he said the bridge was 230 feet high. Let's call that 225,
-> which is 15^2. So, neglecting air resistance, the body would hit the
-> water at \sqrt(64 x 225) or 120 feet per second. That's about 83 miles
-> per hour - considerably less than the terminal velocity for a person
-> without a parachute. Since air resistance is roughly proportional to
-> the square of velocity, even at the end of the fall the resistance
-> would be only a small fraction of the weight. For most of the fall, it
-> would be only a *very* small fraction.
->
-> In this case, air resistance would slow the fall only slightly.
->
-> dow

For the heck of it, I wrote a few lines of BASIC that simulate the
fall, taking air resistance into account. I assumed a terminal velocity
of 200 feet per second, which is about 135 mph. I'll append the code
below. The result is that the body hits the water at 111 ft/sec, or 76
miles per hour.

dow

---------------------------------------------------

' bfall
g = 32 ' ft/sec^2
h = 230' bridge height
v = ' speed of fall
s = ' distance fallen
t = 200' terminal velocity, ft/secdt = .001' iteration time interval, secs
DO
v = v + g * dt * (1 - (v / t) ^ 2)s = s + v * dt
LOOP UNTIL s >= h
PRINT v; "ft/sec, or";
PRINT v * 3600 / 5280; "miles per hour"
END

----------------------------------------------------

David Williams

Apr23-04, 03:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>-> David Williams wrote:\n\n-> > If there\'s no air resistance (which a parachute would magnify), the\n-> > formula you want is:\n-> >\n-> > v^2 = 2 x g x h\n-> >\n-> > where v is the speed at which the body hits the water, g is the\n-> > acceleration due to gravity, and h is the height of the bridge. In the\n-> > kinds of units you appear to like, g is 32 ft/second^2, h would be in\n-> > feet, and v in feet per second.\n-> >\n-> > Any real person would be somewhat slowed down by air resistance, but\n-> > not much, without a parachute.\n\n-> Well air resistance is a huge factor. The terminal velocity for a\n-> falling person is somewhere in the range 120-150 mph. A practiced\n-> skydiver can streamline his profile and achieve up to 180 mph.\n\n-> -drl\n\nI think he said the bridge was 230 feet high. Let\'s call that 225,\nwhich is 15^2. So, neglecting air resistance, the body would hit the\nwater at sqrt(64 x 225) or 120 feet per second. That\'s about 83 miles\nper hour - considerably less than the terminal velocity for a person\nwithout a parachute. Since air resistance is roughly proportional to\nthe square of velocity, even at the end of the fall the resistance\nwould be only a small fraction of the weight. For most of the fall, it\nwould be only a *very* small fraction.\n\nIn this case, air resistance would slow the fall only slightly.\n\ndow\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>-> David Williams wrote:

-> >[/itex] If there's no air resistance (which a parachute would magnify), the
-> > formula you want is:
-> >-> > v^2 = 2 x g x h-> >-> > where v is the speed at which the body hits the water, g is the
-> > acceleration due to gravity, and h is the height of the bridge. In the
-> > kinds of units you appear to like, g is 32 ft/second^2, h would be in
-> > feet, and v in feet per second.
-> >-> > Any real person would be somewhat slowed down by air resistance, but
-> > not much, without a parachute.

-> Well air resistance is a huge factor. The terminal velocity for a
-> falling person is somewhere in the range 120-150 mph. A practiced
-> skydiver can streamline his profile and achieve up to 180 mph.

[itex]-> -drl

I think he said the bridge was 230 feet high. Let's call that 225,
which is 15^2. So, neglecting air resistance, the body would hit the
water at \sqrt(64 x 225) or 120 feet per second. That's about 83 miles
per hour - considerably less than the terminal velocity for a person
without a parachute. Since air resistance is roughly proportional to
the square of velocity, even at the end of the fall the resistance
would be only a small fraction of the weight. For most of the fall, it
would be only a *very* small fraction.

In this case, air resistance would slow the fall only slightly.

dow