Simple problem

[dirtyamerica]

Hey, I've got a question I've been thinking about. I'm going on a rafting trip next month. There's a rope swing next to the river. If I want to swing on it, let go and get the most distance at what point in the arc of the swing do I let go? I realize that my speed is greatest at the bottom...and to release when traveling at 45 degrees with reference to the water (normally the best trajectory) I will almost be stopping at that point in the arc. Please help so I can impress people with my distance, hehe

[5614]

As a guess I'd say somewhere between the bottom and 45 degrees.

 

Dunno, see what others have to say too.

[Cap'n Refsmmat]

I've read that 35 degrees is the best trajectory if you're going for distance, as it also compensates (somewhat) for wind resistance. Although I doubt you'll have a protractor with you to estimate where you should let go.

[insane_alien]

id guess 30 degrees. on practical experience

[dirtyamerica]

As a wild guess, I was thinking 22.5 degrees as it's a comprimise between the trajectory angle and speed. But I really don't know, hehe. It's fun to think about though. And it will be difficult to know the exact angle without said protractor and I'll be drinking too.

[5614]

Yeah. I'd like to see you let go at exactly 22.5 degrees!

[swansont]

It's not going to be a fixed angle, I think. You have to solve in terms of the initial angle, since you can't exceed that, and it also depends on how far you drop.

[J.C.MacSwell]

Exactly at the bottom if you will "land" after infinite drop and exactly at 45 if you have to "land" at that height. (the 45 degree height)

 

So not enough info.

[Rocket Man]

it'll depend on the initial height/starting angle.

there will be a balance between the parabolic trajectory and the velocity at release.

(should this go in the mathematics forum?)

i'll give it some thought

[Rocket Man]

ok, this doesnt account for height above the water on release, but it works none the less.

i went from the trajectory being

s = sin(theta)*cos(theta)*v^2 *2*(1/g)

where theta is the angle of release measured from vertical

and

v =( (sin(phi)-sin(theta)) * r * 2g)^.5

where phi is the starting angle from vertical and r is the rope length.

 

i simplified it down to,

s = sin(theta)*cos(theta)*(sin(phi)-sin(theta))*r*4*(1/a)

(bit of a mess i know)

 

when you start with a horisontal rope, 25.2 degrees is your best bet

when you start at 45 degrees, 19 degrees is best