silly question but ...

scott_e · Post by **scott_e** » Tue Mar 20, 2007 5:21 pm

how fast would an elise need to move in order for someone to jump clean over it ? .... its pretty low so im sure its possible.

anyone tried

skellyjohn · Post by **skellyjohn** » Tue Mar 20, 2007 5:24 pm

I'll drive, you jump!

scott_e · Post by **scott_e** » Tue Mar 20, 2007 5:27 pm

i asked for that i guess

went out for a walk at lunchtime and had to cross a busy road ... a car got too close and the question popped into my head.

robin · Post by **robin** » Tue Mar 20, 2007 5:45 pm

See Newton ...

How high can you jump, and will you be tucking your feet in once in flight or will you be keeping legs stretched out behind you (i.e. as they would be just at the moment you took off)?

Cheers,
Robin

scott_e · Post by **scott_e** » Tue Mar 20, 2007 5:53 pm

robin wrote:See Newton ...How high can you jump,

mmm ... getting old , legs kicked to bits from footie ... probably not high enough these days.

PhilA · Post by **PhilA** » Tue Mar 20, 2007 6:09 pm

scott, ur tiny, so just duck

scott_e · Post by **scott_e** » Tue Mar 20, 2007 6:22 pm

PhilA wrote:scott, ur tiny, so just duck

.... at least i will fit in a lotus Phil

ExigeKen · Post by **ExigeKen** » Tue Mar 20, 2007 7:27 pm

scott_e wrote:
PhilA wrote:scott, ur tiny, so just duck
.... at least i will fit in a lotus Phil

Bitchy

Post by **pete** » Tue Mar 20, 2007 10:00 pm

Like this?

http://gorillamask.net/stunt.shtml

james · Post by **james** » Wed Mar 21, 2007 11:59 pm

Measure how long you can jump up and stay higher than the highest point of the car. Lets say .5 second estimated - because you wont be high enough for much longer I guess.

Take that time and measure the length of the car (lets say 3m) . Add a bit for reaction time etc. (So lets say 5m)

Speed = Distance/Time

= 5m/.5 secs
= 10m per second
= 22mph

Anyone remember more about physics class than me?

robin · Post by **robin** » Thu Mar 22, 2007 9:03 am

s = ut + 0.5at^2

u = (your starting velocity, i.e. your speed just as you leave the ground)

a = -9.8m/s/s (otherwise known as g).

t = elapsed time from when you left the ground.

s = distance from the ground at time t.

Also remember that your average human might be able to do work at 500W absolute peak power, this work being used to give you kinetic energy which is:

0.5*mv^2

Assuming you have a mass of 80kg (fat boys like me don't need to try this!) and it takes you no more than 1 second to leave the ground then your maximum kinetic energy is going to be 1 * 500 and so

v = sqrt(500/40) = 3.5m/s

Let's say with some really hard work, good technique, etc., you might be able to make 5m/s.

Now you can solve the next part, because you can plug in 5 for u and then you'll see that there are two values of t for which s is 1m - one is about 0.25s and the other about 0.75s.

Hence James' guess was just about spot on at 0.5s!

Assuming you synchronise perfectly, the car needs to travel it's full length in 0.5s, i.e. 6m/s or 13.5mph.

For you to clear it you need to leave the ground when the car is 0.25s away, i.e. 1.5m.

BUT you need to work on your technique, because at 4m/s take off speed, your feet won't clear the car at all

You do have the option of tucking your feet in, which I think is a must, that let's you cheat by reducing your overall speed but increasing the speed of your feet/legs. Remember, your head and arms are already high enough in the air to miss the car, so no need for you to raise them up at all!

Cheers,
Robin

james · Post by **james** » Thu Mar 22, 2007 9:53 am

james wrote: Anyone remember more about physics class than me?

That would be Robin then

Rich H · Post by **Rich H** » Thu Mar 22, 2007 10:25 am

That vid demonstrates the technique, except on a liz you would put your foot through the rad, then crash through the soft top and get brained by the roll bar.

I'd probably give it a miss....

jason · Post by **jason** » Thu Mar 22, 2007 11:17 am

Can I be the first to say.... DON'T DO IT!

For some unexplained reason, this has caused me to (think I) remember the same stunt being done by karate guy on that cheesy 80s program "Just Amazing". And the same vague memory is telling me the guy smashed his lower trying it.

They had a long lens camera side on so he could rehearse his jump moment with the car passing him (as opposed to smashing him). He just jumped a little late on the 'real one'.

scott_e · Post by **scott_e** » Thu Mar 22, 2007 4:09 pm

robin wrote:s = ut + 0.5at^2

u = (your starting velocity, i.e. your speed just as you leave the ground)

a = -9.8m/s/s (otherwise known as g).

t = elapsed time from when you left the ground.

s = distance from the ground at time t.

Also remember that your average human might be able to do work at 500W absolute peak power, this work being used to give you kinetic energy which is:

0.5*mv^2

Assuming you have a mass of 80kg (fat boys like me don't need to try this!) and it takes you no more than 1 second to leave the ground then your maximum kinetic energy is going to be 1 * 500 and so

v = sqrt(500/40) = 3.5m/s

Let's say with some really hard work, good technique, etc., you might be able to make 5m/s.

Now you can solve the next part, because you can plug in 5 for u and then you'll see that there are two values of t for which s is 1m - one is about 0.25s and the other about 0.75s.

Hence James' guess was just about spot on at 0.5s!

Assuming you synchronise perfectly, the car needs to travel it's full length in 0.5s, i.e. 6m/s or 13.5mph.

For you to clear it you need to leave the ground when the car is 0.25s away, i.e. 1.5m.

BUT you need to work on your technique, because at 4m/s take off speed, your feet won't clear the car at all You do have the option of tucking your feet in, which I think is a must, that let's you cheat by reducing your overall speed but increasing the speed of your feet/legs. Remember, your head and arms are already high enough in the air to miss the car, so no need for you to raise them up at all!

Cheers,
Robin

brilliant ! .... that raised a smile on an otherwise dull Thursday noon.

Scott