richnfg Posted December 2, 2004 Share Posted December 2, 2004 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! Link to comment Share on other sites More sharing options...
Ophiolite Posted December 2, 2004 Share Posted December 2, 2004 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". Link to comment Share on other sites More sharing options...
Mokele Posted December 2, 2004 Share Posted December 2, 2004 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 Link to comment Share on other sites More sharing options...
richnfg Posted December 2, 2004 Author Share Posted December 2, 2004 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! Link to comment Share on other sites More sharing options...
[Tycho?] Posted December 2, 2004 Share Posted December 2, 2004 Note: these are equations for the man falling. If he jumps upward, it would be a different number. Link to comment Share on other sites More sharing options...
richnfg Posted December 2, 2004 Author Share Posted December 2, 2004 Yep thanks! Link to comment Share on other sites More sharing options...
5614 Posted December 2, 2004 Share Posted December 2, 2004 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. Link to comment Share on other sites More sharing options...
swansont Posted December 2, 2004 Share Posted December 2, 2004 '']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. Link to comment Share on other sites More sharing options...
richnfg Posted December 3, 2004 Author Share Posted December 3, 2004 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. Link to comment Share on other sites More sharing options...
Babbler Posted December 5, 2004 Share Posted December 5, 2004 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? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now