Quick Question

[richnfg]

Is there a formula to work out the speed in this situation?

 

A 65kg man jumps off a 2m high wall, could I find the speed from this (by time he hits the ground)?

 

Thanks!

[Ophiolite]

v^2= u^2 + 2a.s.

 

Where v= final velocity in metres/sec, u=initial velocity in metres/sec, a= acceleration due to gravity in metres/sec/sec and s= distance in metres.

 

 

Edit: I'm curious as to why you didn't just google "equations of motion".

[Mokele]

Just use kinetic and potential energy. Potential energy = mgh, kinetic = 0.5mv^2 (that's supposed to be "v squared")

 

When on top of the wall, the man has all potential energy, and no kinetic. When at the bottom, he has all kinetic and no potential. Since we know the total energy must always be same, we know the initial potential energy must equal the final kinetic energy. So:

 

mgh = 0.5mv^2

 

Then cancell out the mass,

 

gh = 0.5v^2

 

Multiply by 2, and square root each side

 

sqrt(2gh)=v

 

So the sqaure root of 2 times the gravitation acceleration times the height equals the final velocity.

 

 

Alternatively, you could be a smartass and say "0, he stops when he hits the ground."

 

Mokele

[richnfg]

I'm sorry, I could of searched Google. I needed a quick answer as I need it for my presentation tomorrow and I'm in a rush!

[[Tycho?]]

Note: these are equations for the man falling. If he jumps upward, it would be a different number.

[richnfg]

Yep thanks!

[5614]

im guessing that this is related to this:

http://www.scienceforums.net/forums/showthread.php?t=7691

thread.

 

saying that as it is a follow on from this thread and expansion on detail and all.

[swansont]

'']Note: these are equations for the man falling. If he jumps upward, it would be a different number.

 

They work for a man jumping up, too.

 

What it doesn't do is tell you about vector information if he jumps forward as well as falling or jumping up - you'll get the impact speed, not the vertical speed. And it ignores air resistance.

[richnfg]

im guessing that this is related to this:

http://www.scienceforums.net/forums/showthread.php?t=7691

thread.

 

saying that as it is a follow on from this thread and expansion on detail and all.

 

Yeah! I was going to post them in the same bit but I didn't.

[Babbler]

I'm sorry, I could of searched Google. I needed a quick answer as I need it for my presentation tomorrow and I'm in a rush!

 

Your aren't going use it for practital use, are you?