Looking for the math to solve a friendly debate.

We had 5 people hanging out at a pool at a hotel. We started talking about whether it was possible or not to make the jump from a 2nd floor railing into the pool. 3 of the 5 of people don't think the jumper would make the pool, and 2 think the jumper would make it.

Here are the numbers:
* 2nd floor rail is 14 feet high. (person would be standing on top of the rail)
* Concrete clearance from rail out to edge of pool is 11 feet.
* Requires a standing jump. No running start.

* Person is able to make a standing jump from edge of pool into pool of 8 feet. This jump was a relatively flat angled jump (we don't have the exact angle). For this test, the edge of the pool is about 1 foot above surface of the water if that factors in.
* Person is 5'10" and about 180pounds.

The debate:
Can this person make a standing jump from the top of the rail on the 2nd floor and clear the concrete on the first floor enough to land safely into the pool. So the person would need to clear more than 11 feet as that is just to the very edge of the pool/concrete.


Disclaimer:
This will never be tried. Just a friendly debate if the person would be able to make it or not


If the jumper can jump 8 feet out, before he starts falling below the 14 foot height mark, how many more feet will he clear in the 14 foot drop before hitting the concrete/pool?

What other information would be needed to solve this mathematically?

The person's trajectory is modeled by a parabola. Knowing that he can make it 8 feet out means that you can model his jump as f(x) = -A*x(x-8)+14. The value of A is the important value here.

You want it to be the case that when x = 11 (when he's 11 horizontal feet from the launch position) that his height is positive: f(11) = -33A + 14 > 0. So A < 14/33 = 0.4242.

You either need the launch speed or launch angle to determine A. This can be determined by knowing how long he's in the air for the 8 foot jump. And that's probably the easiest way.

Some practical considerations also include things body position in the air. His center of mass may travel in a parabola, but if his body is positioned the wrong way he would have his center of mass clear the concrete but his body won't.

When a reasonably fit person jumps for a basketball jump-ball situation or any jump from standing position with maximum effort placed they typically will not make more than 1m elevation in their center of mass. One can even argue that maybe 0.8m is more reasonable.

That would suggest the jump velocity from rest if standing on the edge is 1/2*mv^2=mgh or v=(2*g*h)^(1/2)=3.96m/sec

So lets say typically one can start from rest and assume they can accelerate during a bending and release of legs to full extension from rest without any prior running, just standing at the edge there (supported by some method to not drop prematurely) and they jump with immense effort delivering 4m/sec even 3.5m/sec for safety.

In that case the lowest part of the body say starts from height 4.27m and has to clear 3.35m plus room for error.

To leave some room for error ideally one wants to clear the point 3.35+0.3, -4.27 if one is jumping from what we view as origin at 0,0.

To achieve that you need to use the trajectory equation (that describes all the points x,y that this curve will ever pass from provided it doesnt hit any obstacles on the way to them)

https://en.wikipedia.org/wiki/Projectile_motion

where one has expressed y as function of x using as parameters the tangent of the angle of launch θ and the initial jump speed u.

y=Tan(θ)*x-1/2*g*(1+(Tan(θ))^2)/v^2*x^2

Your point of clearance is say x=3.35+0.3=3.65, y=-4.27 (since gravity is significant on earth the dominant velocity component at the edge of the pool is the vertical one for most possible by humans orbits , so i only need to worry about a horizontal clearance issue)

You need to find the set of u,θ that allow us to reach that point or better.

The above equation is a quadratic in tan(θ) and must have real solutions for u if reaching the x,y point is to happen. That will give you the condition that u,θ must satisfy.

As one can show easily using the discriminant of the tan(θ) quadratic (or see the above link under the section "Angle θ required to hit coordinate (x,y)")

u^4-g*(g*x^2+2y*u^2)>=0

this is the same as u^4-9.81*(9.81*3.65^2+2*(-4.27)*u^2)>=0

this has solution u>3.64m/2

But it gives very little room for error in the launch angle.

A safer speed like 4m/sec (but questionable whether achievable by most humans) would allow θ to range from 0 to 43 deg say that is doable as aim.

First of all i say immediately this is a no way reasonable jump because you can die or become paralyzed with some significant probability if you perform it and hit the floor or the edges.

To know if one can do it if you have to (without initial running speed possible due to the balcony geometry), just stand at the edge of the pool and jump with all your strength and try to launch with angle say 20-45 deg and so jump and see how far into the water you land. Take a video and examine the orbit.

The range is u^2/g*sin(2θ) so if for 20 deg you can land at least 1 m into the water then you have near 4m/sec. To be safe try to go for 1.2 1.3 m with a few test jumps. If you have that then maybe you can do it.

Air resistance will inhibit the actual risky jump a little but i doubt its significant at such low x component speeds. You will be in the air for about a second it seems and the x position remains close to the no air resistance solution.


Try this to see what is going on

https://www.desmos.com/calculator/reesidrq8h


To jump with such speed as near 4m/sec and such small angle you need to have some support in your launch segment that allows you to extend and accelerate in a direction that is far from vertical that may make it harder. Just do a test jump as i suggested from floor level and see how far you land. If you cant clear 1.1-1.2m its probably bad. Without any support in place if one stood up there in the balcony edge and jumped, there is such a significant risk something can go wrong, that its a very bad idea. At the very least if one dared to do it put multiple mattress type protective material (eg some of them used in gyms) all the way to the pool edge.

If you land bad you will at least not die but can still get injured even with the "mattress".


Ps:If you have to do it because your life depends on it i bet you can clear it.

If the guy can clear 8ft or 2.4m on a floor to pool jump then he can easily do it from the rails if he does the same thing ignoring the fact its so high because he likely has over 4-4.5m/sec jump speed. But speaking of a random person the proper thing is to first try the test jump from floor level and make sure you have at least 1.2m range. It seems doable by most but still not a good idea because something can go wrong while attempting it.

The rails inhibit a little the jump because in the floor example your feet is perfectly launched from a totally big flat area and the same is not true from the rails. I would imagine you need your entire sole or the very ends of the sole near the toes being firmly stable during acceleration and deriving reaction force from as much as possible area during the entire launch sequence.

So to make a realistic jump try to test jumping from something as wide as the rails like a set of tiles inhibiting the sole from having full access to the floor or put something on the rails like a platform. Standing on the rails and quickly jumping without any support or preparation during the launch sequence is exceptionally risky because you may never have access to the full proper acceleration sequence that you enjoy from the totally comfortable wide flat surface. If you drop from 4.5 to 3.5 m/sec its game over. Also when jumping from a height and trying to maximize range the solution is always for a given speed an angle less than 45 deg.

First, let me be clear. This will NEVER be attempted (by any of us at least)!!! This was just a debate whether or not it could be done.


The rail is actually pretty wide, so the sample jump from the edge of the pool into the pool seems like a good test. That jump was a standing jump from the edge of the pool and the first point of contact in the water was 8 feet from the edge of the pool where the jump started. The edge of the pool was about 1 foot above the surface of the water so that plays slightly into it.


If I had to guess the angle of the test jump from the edge into the pool, I'd say about 20-25 degrees.


To be perfectly honest, this math mentioned in the above threads is above my head . I was hoping for someone to be able to use the math above and say...yes, he could make it easily, or no way.


Is it somewhere in between an obvious yes or no?

All you need is to play around with my Desmos link and see what happens for various speeds and launch angles (angles in radians to get to degrees multiply by ~57.3).

Here is the frozen trajectory that fits the orbit the guy would have had if he launched as he did from floor level. The horizontal line is to show this as reference and to also use as reference for anyone testing in advance of trying the real thing.

If you hit play on the a (angle) and u (speed) parameters in SI units you will see what happens for a number of other choices of orbit parameters with the obvious points of launch and safe landing indicated with dots (i have him 0.3 m room past the edge of the pool just to be a little safer in clearance.

https://www.desmos.com/calculator/eoio3ons0x


As you see if i take your word for his range and the launch angle near 20 deg i get speed 5.3m/sec that i find a little too optimistic for most people. He easily clears it then.

idiots...

Speaking of jumping properly.

From Greece with love!




She is good and beautiful. Nothing wrong with being a little sexy too.


And this is the jump that haunts us in history because its probably one of the most difficult records to break;

https://en.wikipedia.org/wiki/Men%27...rd_progression

The most unreal day ever in the sport (1991). Because the previous record was from 1968. Today 2019 we are further away from it that it was from the previous holder. That is the definition of tough.

And a little coolness in action courtesy of my good friend BruceZ

Cheers



We should make that the official video of the life is drunk thread.

https://www.usna.edu/Users/physics/m...Projectile.pdf

I was thinking MythBusters... throw a crash dummy over simulating a jump and see what happens. It's both physics/engineering... are you really going to go through the math and jump yourself? I doubt it. You may as well get a crash dummy.

I don't know what most of those words mean. I only know that the distance gravity makes you fall (or jump up) is 16 t squared. A guy who reaches a height of two feet will take about .35 seconds to fall back and about .35 seconds to reach that height. So his horizontal velocity is almost 12 ft per second if that jump gets him a distance of 8 ft. (assuming no air resistance and a straight leg landing.) The same exact jump from 14 ft up takes 1.35 seconds so with no air friction he easily makes it. In real life its much closer because he probably ended his initial jump with bent legs and /or his vertical jump was a bit higher and there is air resistance.

My objective as always is to find more than what is asked like what is the minimum speed that for some angle will make it or to be able to optimize and find the best angle or be realistic and assume humans cannot be that accurate and there is also air resistance but not yet important at such speeds (terminal velocity for humans is like 54 m/sec or ~120 miles per hour so at only 1/10th of that horizontally it will not matter and vertically it will even help by making it longer to fall by a tiny bit. (i use SI units by the way because all the scientific community planet does that because its idiotic to use units that have feet, yards, inches and miles all with strange ratio numbers connecting each other having to remember endless rules like 1 inch is 2.54 cm and 12 inches is 1 feet and 3 ft is 1 yard and 1 mile is 5280 ft etc all madness. In these units g = 9.81m/sec^2 or for easy calcs 10m/sec^2 )

You do know what i am talking about. You cam recognize very well that horizontally ignoring air resistance there is no acceleration so distance is proportional to time and the rate is the horizontal component (cosθ) of the velocity vector at the start u0 and that vertically gravity exists to create a constant acceleration problem

so x=u0*cosθ*t, y=y0+u0*sinθ*t-1/2*g*t^2 (E1)


So of course if one only wants to get a crude idea with even as little as 4 m/sec and a small angle like 20-30 deg (must be less than 45 when trying to go lower than horizontal level in range) the x component of velocity is 4*cos25~3.63 m/sec and even if one only slightly estimated the time used to go up to 0 vertical velocity ie 4*sin25/9.81=0.17sec and assumed vertical drop with height 4.27m plus a bit more than that time, they would get total flight time a little more than (2*4.27/9.81)^(1/2)~0.93+0.17~1.1 sec during which x will advance from above by more than 3.63*1.1~4m that is beyond the needed.

I have to play that way and be conservative in estimates and even cut a little the total time because i have no idea one can clear the jump with perfectly executed angle like doing it on floor level where the ground and water nearby immediately offers a sense of level direction preventing even very flat launches and because air resistance will add a bit of drop in horizontal speed. From the rail one can even risk losing balance and ending up jumping with near zero vertical speed on the way down last moment managing to accelerate near horizontally.

So with even more speed even safer clearance is possible, like his friend's claim ( that i am not personally buying is true or can be executed that well from the rails with the feet not firmly on the ground and the stress of falling in place).


When we answer problems we care to expand as much as possible and cover all risks otherwise all it takes is to run my desmos simulation.

My point is if you wanted to do a thorough analysis you can do it with what you know and the equations E1 so of course you understand what i am talking about. The objective was to find a realistically minimum and safe (ie big angle range) set of initial conditions that clears it.


The main risk here is jumping with near 0 angle or very high over 45 deg for some reason of not properly executing the launch sequence like she did in that terrible video

Wrong. While the perfect projectile is modelled by a parabola, when you take drag into account, real life projectiles are more accurately modeled by the hyperbolic cosine multiplied by a constant. It is close to a parabola, but it is not a parabola.

The trajectory is not a parabola something known even in 15th 16th century but only if you are considering a very long flight and high speeds and/or small mass objects.


I can show you why it wont matter a lot here so the parabolic treatment is not bad in a significant manner.

If we model the air resistance at the speeds we are dealing with here that remain from 4 to 10 m/sec roughly and the x component about 4m/sec say with F=-kv^2 (instead of -kv at smaller Reynold numbers)

Then for a human that has terminal velocity 180km/h or 50 m/sec and mass say 80kg

we must have mg=k*vterm^2 or k~0.31 Nt*sec^2/m^2 (in SI units for all involved).

So in the x direction the force would be say order k*4^2 or ~ 5N that for a mass of 80 means an x acceleration of -0.0625m/s^2 . So since the entire flight takes a little over a second you do not expect the speed in x direction to change by more than 1.6%. That means even less relative change for the range because clearly the effect is gradual to 1.6% for speed and is much less earlier in the orbit. Additionally the time to fall is increased a bit and that compensates further reducing the correction to unimportant levels for that air resistance profile for a human (for a baseball ball its substantially different story).

By the way this is an approximation for demonstration of size influence, the actual proper calculation of x component deceleration will require to use the total speed and then expand the force kv^2 air resistance force that is tangential and opposite to trajectory into x component the proper way (something that changes over the flight) but it wont matter a lot if done super accurately.

Lol dude, I was trolling Aaron, coz I don't like him. Of course it's a parabola.

kcoshx is how a chain hangs, not how a projectile flies, and it's called a catenary.

Yeah but can you prove it by producing a variational calculus minimization of the potential energy integral (amazon interview on that by the way with a stupid trick instead of the proper solution)?

No. I'm not that good. But Wikipedia can.

No they dont have it proven that way. That is a claim of mine and i think its doable easily with Euler Lagrange equation on the potential energy integral using arc length ideas to represent the potential energy integration etc.

The curve in the air resistance problem for a human is definitely not a parabola to be super accurate but its not very bad to imagine it is as long as speeds remain sub 10 m/sec here and flights order 1 second.

So if the person jumps and marginally fails, it is definitely air resistance that did it but probably its not altering more than 10cm the whole range thing.

Erm, you forgot the adjustments for special and general relativity.

No the air resistance correction here is real not tiny and academic but not super material to enter the discussion hence the suggestion for having over 4m/sec. There will be order 10 cm difference and a lot more if the effect involves bigger speeds and heights or sports balls etc.

Hang on, you figured out a new equation for the catenary? Do tell.

Take the potential energy integral over a proposed y(x) curve line.

U=Integral ( g*dm*y(x))=Integral ( g*ds*L/m*y(x))=Integral ( g*(1+y'(x)^2)^(1/2)*L/m*y(x)dx) and now apply Euler Lagrange equations on

F(x,y(x),y'(x)) =g*L/m*(1+y'(x)^2)^(1/2)*y(x)


ACosh(B*x) verifies the resulting differential equation.

I know the Amazon question, by the way. The poles overlap. You saw that on youtube, Presh Talwaker's channel.

I understood about 1 out of 3 words in this thread, but got the message that it seems math supports the idea that the jumper would in fact make the pool. Only thing people are still trying to figure out is how easily he'll make it.

Thanks all!!!