Physics Help!! Critical angle of Trajectory

[Challenger]

Well me and about a 1/3 of my class have been working on this problem for our takehome, been at it for 4 hours.

 

The question is what is the critical angle to the horizontal at which a stone can be thrown and always be moving away from the thrower. This means the distance from the origin to any point on the trajectory will always be increasing. We need to find the angle theta where this becomes true.

 

Weve been working on all sorts of ideas, none seem to work but with a little internet searching Ive found some people saying its about 70.83 degrees but there is no proof, what we need. The tests due tomorrow at 9:30 so we really need this soon its worth alot of points!!

 

Please any help is appreciated, I know theres alot of engineers here, someone must know or have ideas.

 

Thanks

[m4xwellmurd3r]

umm...ANY angle you can throw a stone and it'll move away from the thrower unless you threw it perfectly at a 90* angle that is. think about it. throw it at 1 degree from horizontal, it goes forwards, throw it at 89 degrees from horizontal, it moves forwards. the stone makes an inverse parabola no matter what the angle. hell you could throw the stone at -89* from horizontal and which way will it move? away from you.

 

common sense man

[Challenger]

No we are trying to find at what angle it will always be increasing and getting further away, it cant go away from you and then start to get a bit closer, something you get with an angle close to 90. When you have a angle closer to 0 the distance from the origin to a point on the trajectory will always be increasing.

 

Its not how you describe it.

[m4xwellmurd3r]

unless you have wind it will never move close to you at an angle greater than -90 or less than 90 it's IMPOSSIBLE to do that without wind. at 90* it moves straight up, and lands on your head. at -90 it moves straight down between your feet. at 89 it moves slightly forwards from your hand, and lands forwards of your hand, at -89 it moves straight down, a bit in front of you. UNLESS you have wind, it can't possibly move BACKWARDS if you throw it FORWARDS. so unless there's an outside force working against that stone, it moves forwards.

 

mythbusters had this slight problem when they did the bullet shot in the air trick. they had to get a PERFECT 90* trajectory to get the bullet to stop in the air and start tumbling at it's max velocity of 120mph into the ground (even then it moved away from the gun many feet)

 

it's a trick question of sorts. you can't throw something forwards and have it come back to you if you don't have any wind. that's just physically impossible. OH, and you can't count on it BOUNCING backwards. That wouldn't count, since as soon as it hits the ground, THAT is the landing point. go outside and get a sling shot, shoot it straight up in the air, and I'll bet if you shot it right, no matter what angle you do (as long as there's no wind) it'll never get closer from your shot point

[Challenger]

If you throw something at about 75 degrees the distance you are away from it (distance from origin to the point ON the trajectory path) at the peak height will be less than the distance from the origin to the landing point.

[m4xwellmurd3r]

so what you're asking is slightly different eh? you need to have the distance between the PEAK and the landing, be equal or greater than the distance between PEAK and origin?

[Challenger]

No.

 

We are trying to find the angle of launch that the distance between (0,0), the origin, and any point on the trajectory will always increase

 

I was using the example of 75 degrees to show that the distance between you the thrower (origin) and the rock on its trajectory path can increase and then decrease. (since the distance from you to the ball at the top is more than the distance between you and the landing spot)

 

Im not talking about x or y distances but the hypotenuse of those, the square root of x^2 + y^2.

 

You get it now?

[xxjoeyxxeb]

Stand on top of a skyscraper and drop the rock(-90deg). Always increasing in distance away from you, or I guess Grand Canyon..

[Challenger]

Between 0 and 90. There is an angle at which the distance I speak of is always increasing. 0

 

Anything angle less then that theta angle the distance will always be increasing, anything more then theta and the distance will increase/decrease.

 

 

The distance from the origin to the top of the tallest trajectory is more than the distance from the origin to the landing spot. Then if you look at the trajectories at smaller angles, the distance from the origin to a point on the blue line is always increasing.

 

We need to find the angle that this becomes true. It seems logical for the angle to be in the 45-70 degree range, but we have no way to find the actual critical angle theta.

[m4xwellmurd3r]

hmm i would think 45 would be the optimal angle, but that's just due to me playing to many trajectory/shooting games.

 

I thougth about it more. 60* would be the limit I think. if you throw it at 60* it will land at 60* (not counting wind resistance) meaning the peak would HAVE to be at 60* hypotenuse would equal distance thrown. throw it at 61* and it lands at 61* the peak has to be 58* meaning the hypotenuse is now greater than distance thrown.

 

I think...

[Challenger]

Yeah almost everyone thought it was 45 at first because it seemed so simple but we actually endeed up getting an answer that we think is right and its about 63 degrees.

 

It had to do with a semi circle with radius r which is the x length of the trajectory. We had to fit a trajectory like the above blue lines in side that semicircle with only one point touching, the end impact point. If the trajectory went outside the semicircle then we know that it was a longer distance away from the origin so it had to decrease back to r the x length travelled.... so we had a system of equation for a hyperbola, took the derivative solved at 0, found a ratio of sides arctan and we got an angle... hopefully its right..

[blaze73]

Ok so I drew this up in solidworks so I could play with different scenarios. Hopefully the picture helps to show that any smaller of an angle and you wouldn't be maximizing, and any larger of an angle and you will get a situation where the distance increases, then decreases, then increases again. The angle isn't exact but it should be very close (and very similar to that 70.83 number you had before).

 

 

The key is to find x and y such that the functions f(x)=a(x-h)^2+k and g(x)=sqrt(r-x^2) (parabola and circle) equal each other AND the derivatives equal each other (are tangent at that point, same slope). You can pick any value for h and note that f(0)=0, and r=sqrt(x^2+y^2). Use all this info to solve for k because for your initial angle tangent line (y=mx), y=2k at x=h. You can then use arctan(2k/h) to get your angle.

[woldson]

I was thinking straight down starting at your feet into a shallow hole.

You could also throw it straight 0 deg. and low on the horizontal axis, like a bowling ball.

[Challenger]

The question wasnt what is "a" angle, it was what is the critcal angle where it goes from always increasing to increasing/decreasing.