Donate SIGN UP

the great escape

Avatar Image
gontofe | 16:39 Wed 17th Sep 2003 | Travel
8 Answers
Imagine (this may be hard) that I'm in Essex in my Fiesta 1.1LX. I want to leave badly, but don't have money for the dartford crossing, so I decide that (with certain modifications) I want to jump the Queen Elizabeth II bridge to miss the tolls. How fast would I have to be travelling (I realise my car doesn't go this fast, but let's say it's powered by a jet engine)?

Answers

1 to 8 of 8rss feed

Best Answer

No best answer has yet been selected by gontofe. Once a best answer has been selected, it will be shown here.

For more on marking an answer as the "Best Answer", please visit our FAQ.
-- answer removed --
Question Author
...and just when I thought that (from what I've been reading) that AB was going to pot, and that there was a lot of sniping and backstabbing going on, finally along comes someone with a sensible answer! Thanx!
-- answer removed --
-- answer removed --
If you can't afford the �1 then you couldn't afford the rocket fuel. My advice would be go up the A13 into london then through the blackwall tunnel.
As the QEII bridge is 2.8 kilometres long according to this website http://www.dartfordarchive.org.uk/technology/commu
nication_drc.shtml
and taking your fiestas wieght as a standard 874 kilos (you don't specify the year so I'll have to guess at this one) then with a take off ramp at an angle of say 45� then you can use he equations on the following pages to work it out for yourself....

http://www.usafa.af.mil/dfp/cockpit-phys/md2th2.ht
m

http://webphysics.iupui.edu/152/152sp03/152Basics/
projectiles/projectiles.html

http://www3.district125.k12.il.us/faculty/tgraba/A
ccelerated/projmotioneqs.html

I'm not going to do all the hard work for you....Good Luck

Question Author
Wow! Now *that's* an answer!
You got me to thinking though and I tried a few figures out....It's really not an easy thing to work out exactly as physics equations tend to ignore things like wind resistance and other variables and only include constants like gravity (9.81 m/s/s) but i came up with a rough estimate of somewhere in the region of 420-450 miles per hour for a take off angle of 45 degrees....I stand to be corrected on this by anyone with a better working knowledge of the equations though.

1 to 8 of 8rss feed