l0g1c wrote:Btw, I'm still right, but I've only given how to do the last part, you still have to infer what one of the other distances is by noting the 60 degree angle.
Are you taking into concideration the slowdown and increased drop rate at the far end of the path? It wouldn't be symmetrical. Or are gravity and wind resistance not being used? Guess I better read the whole thread. I read some of it last night, but not all of it today.
guys - just read my solution - i used two equations that were posted on a physics page, and solved for the height
you can double check - it's not like you have to trust that i'm right - you can see for yourself whether i'm right or wrong (and i don't purport to be 100 % correct)
Last edited by [xeno]Julios on Sun Apr 03, 2005 9:57 pm, edited 1 time in total.
trajectories are always symmetrical in a constant gravitational environment so long as there is no air resistance and the starting and ending heights are the same.
I took air resistance out of the equation to simplify it. As far as gravity goes, gravity is a constant, so the resulting trajectory will always be symmetrical, right? I'm at work now, so my thinking is a little sporadic. I'll go home and cheat (read Julios' page) and see how off I am.
phantasmagoria wrote:it won't be symmetrical if the projectile is fired from above the horizon..
[xeno]Julios wrote:
trajectories are always symmetrical in a constant gravitational environment so long as there is no air resistance and the starting and ending heights are the same.
Just from skimming this thread, it appears that Jules is most likely correct, although there's too much rust on the portion of my brain that deals with 2D kinematics. I'm buried under an electronics project atm, but I'll sort it out...
BTW Jules -
I was just giving you crap. I didn't really see a problem with your solution per se, I just wanted to see what the response would be if I said I did.
I do that with my students sometimes to see how confident they are in their answer....they hate it.
