Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: ericthesheep on 20/07/2009 03:40:15

I'm on a flat roof 20 feet high. The pool is 20 feet from the house.
Do I make it?? With and without running start, assuming I have 20 feet to get a running start??
One of us thinks that regardless of how high the roof is you can only jump as far as you would if you were jumping from 'ground level'  If I can take a running start from 20 feet and the furthest I'm able to jump is 10 feet then I'll only be able to jump 10 feet from the base of the house??
The other thinks that you can jump considerably further if you jump from a higher level. Perhaps because of trajectory and/or having more time to move forward??
Would this be the same without a running start, and just standing and jumping as far as one can??
Any help with this would certainly be appreciated.
Thanks a lot.

It does depend on your trajectory. Here's a diagram showing three different trajectories, one succeeding and two ending with a trip to the emergency room or worse. The proportions are a bit wrong, but if the house is 20' high and the pool is 20' away, and assuming you do the same 10' jump you did while running, you can see here all three trajectories intersect 10'from the original place, but depending on the angle of attack have different parabolas and success. To get the most distance when aiming down at something its usually a good idea to shoot straight out, not waste any energy on jumping vertically. Anyway this experiment should not be attempted, because there is little chance anyone would be willing to replicate the results, or scrape you off of the deck.
[diagram=484_0]
i cant figure out how to delete the extra lines from the curves, in anycase for the trajectories just pay attention to the curved lines.
also to answer your questions in general, yes, you'll always jump farther from higher if you perform the same exact jump, but by how much is the question you and your buddies should have been asking.

Hmm... I'm not quite sure about that diagram as the blue trajectory, which lands in the pool, appears to be circular.
Ballistics, on Earth at least, is pretty straightforward. The lateral distance you travel just depends upon your lateral speed and the duration of your 'flight', with the duration of your 'flight' depending on how high you jump and how far you fall.
Working it out is pretty easy but you need to know the horizontal and vertical speed vectors at the point that you make the jump. Using your vertical speed you can work out how high you'll jump against gravity and for how long you'll be in the air until you hit the ground and you then just use this duration to work out how far you will have traveled in the horizontal direction, based on your horizontal speed. At the speeds you're likely to achieve by running, your slowdown due to air resistance will be negligible.
A long jumper, of the athletic type, not one of the neilp derived clothing items, sprints down the runway to maximise their horizontal speed and then tries to jump as high as they can.
For the simplest solution i.e. just running off the roof and not jumping up at all:
From s = ut + 1/2 at^2
where s is displacement (20 ft), u is initial speed (in this case 0 because we're just falling and not jumping up first), t is the time and a is acceleration (32.2 ft/s^2):
t^2 = s / (a / 2)
t^2 = 20 / (32.2 /2)
t^2 = 1.242
t= 1.114 seconds
So to travel 20 ft horizontally in a 20 foot drop you'll need to be moving at 20/1.114 feet per second = 17.952 fps = 12.24 mph
That would land you on the edge of the pool though, so you should probably add at least another five feet and use 25/1.114 = 22.441 fps = 15.300 mph.
Have a look at http://en.wikipedia.org/wiki/SUVAT_equations#Relative_velocity (http://en.wikipedia.org/wiki/SUVAT_equations#Relative_velocity) and open up the two examples.
DO NOT TRY THIS EXPERIMENT BASED UPON MY MATHS

DO NOT TRY THIS EXPERIMENT BASED UPON MY MATHS
WHY???

None of those curves look parabolic, aamof. But does it matter?
The optimum launch angle wouldn't be horizontal, in fact. If that applied in general, the max distance over a level surface would be achieved with a horizontal take off. It is actually 45degrees ( if you ignore air resistance) if start and finish are the same height. For the pool situation the optimum will be a bit above horizontal.
The sums are not hard but I'm so racist that I can't bring myself to work in imperial units. I did them at School then we all grew up over here (!!!!!!!) and started using SI.
I think we should do a deal. We (UK) drive on the right and you use SI. How about that?
I wonder who would have more accidents during the changeover.

On a level surface to max out the distance you'd want to shoot at 45°, but that isn't true for shoot down at something. If your target is at a lower elevation, to get max distance, you want to shoot straight out close to 0°.
The curves were the best I could make with the 'make diagram' tool, but all of them intersect at 10' and the diagram was meant to show how different trajectories that result in a 10' horizontal jump, or jump from "ground level". Though they're not parabolic, you'd get a similar diagram on a graphing program, with a higher angle of attack resulting in lower horizontal distance traveled.

How can the optimum 'suddenly' change from 45degrees? The limit, as the height tends to infinity, is zero elevation but you want the right answer for your particular stunt, don't you?
You need to do the sums using the formulae and the differentiate to find the max.

Here's a diagram showing three different trajectories, one succeeding and two ending with a trip to the emergency room or worse. The proportions are a bit wrong, but if the house is 20' high and the pool is 20' away, and assuming you do the same 10' jump you did while running, you can see here all three trajectories intersect 10'from the original place, but depending on the angle of attack have different parabolas and success. To get the most distance when aiming down at something its usually a good idea to shoot straight out, not waste any energy on jumping vertically
I appreciate your enthusiasm for scientific rigor, sophiecentaur, but I did say pretty much what you said in my original post, quoted above.
You, however, said in your original post:
The optimum launch angle wouldn't be horizontal, in fact. If that applied in general, the max distance over a level surface would be achieved with a horizontal take off. It is actually 45degrees ( if you ignore air resistance) if start and finish are the same height. For the pool situation the optimum will be a bit above horizontal.
If "It is actually 45degrees" in general, rockets including the space shuttle, most basketball shots, and artillery canons round the world would be locked in at 45°. And for the optimum solution being a bit above horizontal, that's what my 3 minute diagram shows. The optimum horizontal distance without a ground to hit is found at 0° with a set velocity, just as the optimum vertical distance without a ceiling to hit is found at 90° with a set velocity. We can both agree that the optimum angle of attack for the current situation is not 45°, and is not exactly 0°, but much closer to the latter.
You would need to do the sums and formulae to evaluate the actual distance traveled, you don't need the sums and formulae if you have the principles on hand or in your pocket to shed some light on a drunk argument. Unless of course he gave us some sigfigs. And his calves, roof, concrete, pool, beer, and measuring tape had standard error ratings. Peace?

DO NOT TRY THIS EXPERIMENT BASED UPON MY MATHS
WHY???
Because the 'experiment' proposed by the questioner involved a person jumping off a twenty foot high building. If someone wants to risk their life they should do their own maths.

Pax, tychobrahe, my friend.
I shall still have to work it out in a few spare moments. I remember the 45 deg answer was fairly triv (in the VI form).

The optimum launch angle wouldn't be horizontal, in fact. If that applied in general, the max distance over a level surface would be achieved with a horizontal take off. It is actually 45degrees ( if you ignore air resistance) if start and finish are the same height. For the pool situation the optimum will be a bit above horizontal.
If "It is actually 45degrees" in general, rockets including the space shuttle, most basketball shots, and artillery canons round the world would be locked in at 45°. And for the optimum solution being a bit above horizontal, that's what my 3 minute diagram shows. The optimum horizontal distance without a ground to hit is found at 0° with a set velocity, just as the optimum vertical distance without a ceiling to hit is found at 90° with a set velocity. We can both agree that the optimum angle of attack for the current situation is not 45°, and is not exactly 0°, but much closer to the latter.
The optimum angle for maximum range for a ballistic trajectory is indeed 45 degrees.
Ground launched space rockets and the shuttle don't launch at 45 degrees because they don't follow ballistic trajectories. With a ballistic trajectory the maximum velocity is at the point of launch, as with a cannon shell, but space rockets start at zero velocity and accelerate. Obviously, if you tried to launch a space rocket at 45 degrees it would just fall over.
With basket balls, you don't want a very high horizontal speed when the ball goes through the hoop as there's a risk of it bouncing back off the backboard and missing the hoop entirely. Also, because of the size of the ball and the size of the hoop, there's a minimum angle that the ball can fit through the hoop.
Field cannons are not fixed at 45 degrees because they use shells with fixed amounts of propellant; if they were fixed at 45 degrees they could only ever fire over their maximum range and would be unable to target anything closer. The very large guns fitted to battleships use discrete bags of propellant though, so they'll often be fired at close to 45 degrees to minimise propellant, which in turn reduces barrel wear.
Actually, without working it out, I think that a 45 degree trajectory off the roof would be optimum in terms of requiring the lowest initial velocity.

I think that, to progress this with any authority, someone has to do the actual MATHS. At least, for a human jumping, the speeds are low enough for air resistance to be neglected.
My problem is to decide how to control the take off angle and to decide what it is. Long jump training would teach you the optimum takeoff angle for long jumping but there will be a trade off between velocity, final kick force and direction.
You may well have got me started!

I just found this link (there don't seem to be many of them around)
http://cev.org.br/biblioteca/thebiologicalmechanicsdemonstrationandthepracticalemploymentoftwojumptypetakeoffinlongjump (http://cev.org.br/biblioteca/thebiologicalmechanicsdemonstrationandthepracticalemploymentoftwojumptypetakeoffinlongjump)
One surprising conclusion is that "Reasonable takeoff angle range is 6070 degrees ". Bearing in mind that the jumper's centre of mass ends up lower than it starts, I would have thought they would go below 45 degrees. (By all our previous arguments). That's a bummer.
Couldn't we change the scenario to firing a cannon from a mountain top? It would be much more straightforward. OR, firing our drunken friend / volunteer out of a cannon?

Original poster here....
Thanks for the replies  and I assure you all that the "stunt" was not, and will not be attempted....it was just a friendly discussion between a few halfdrunken buddies who hadn't taken a physics class in about 20 years...
Is it safe to assume that if the jumper did not take a running start from the higher level, and just did a standing jump, that he wouldn't get much farther than if the exact same standing jump would have been performed from ground level??
And if the height were 5x or 10x or 100x as high, I assume that the jumper would eventually stop moving forward and fall straight down?? Regardless of a running or standing jump??
Again, thanks for all of your comments, I appreciate it.
ErictheSheep

The trajectory (without air resistance) would be a parabola which actually would never be exactly vertical. "near as dammit" I hear you cry. Correct.

A straight vertical line would of course be a straight vertical line and not a parabola, but its height charted over time would show that its path was parabolic. And for leeE, a great man once said "Sometimes you have to read between the lines to understand what is being said. Other times, you just have to read all the durn lines." That man was me, just now. I said that those things would be set at 45° optimal angles to show that if a condition were always true it would lead to an absurd conclusion. I already conceded that 45° is the optimal ballistic trajectory for reaching a distance on the same elevation. It is not true for reaching a distance on a different elevation, higher or lower. This thread is done, it just doesn't know it yet.

This link says it all; he's been bothered to do the sums  well done him!
Putting 45degrees into any example gives the max range.
http://www.convertalot.com/ballistic_trajectory_calculator.html (http://www.convertalot.com/ballistic_trajectory_calculator.html)

Heres the numbers. Taking off on level ground at 6m/s at 45° results in a range of 3.66m, or close to 10'. Do the same jump (6m/s) at a height of 6m (around 20') at 45° your range is 6.87m, a significant improvement. However, taking off with a velocity of 6m/s from 6m high at an angle of 26° results in a range of 7.58m, a more significant improvement, especially when your vertebrae are at stake. So, assuming he made an ideal jump at ground level, of 10' range from 45°, with the same power in his legs and with the same running start, the optimal angle from the roof is not 45°. It's close to 26°. The higher you get, the optimal angle for range gets farther and farther from 45° and closer to 0°. 45
Putting 45degrees into any example gives the max range.
This is wrong. I'm just saying. Change the Y_{o} and the optimal angle for range will change. I promise.

Interesting.
My examples were all for higher speeds. As you say, once the speeds get low, the departure from 45 gets more and more noticeable. Well well!
Those 'cheat' numerical methods are great value.
I tried to do it analytically but it ends up with a solution to a quadratic which you then have to differentiate to find a maximum. It's triv for h=0 but for any other value there are too many trig functions and too little time or confidence in my ability to do chains of manipulation.

If the pool is twenty feet away (just to make it easier I'll convert it to meters) That is approximately 6.1m
Gravity pulls you down at a rate of 9.8 m/s^2
If you jump one meter into the air upon departure then you have 7.1 meters until you hit the ground... Giving you .76 seconds before you will hit the ground.
Assuming that you are one meter from the building before you start to drop, you have 5.1m to go in .76 seconds. This would require you to be going at a rate of 6.7 m/s. (even further to make sure you don't hit the edge of the pool)
And 6.7 m/s is equivalent to 22 f/s which equals 15 miles per hour.
That would give you a four minute mile!
I hope you're athletic [;)]

Due to the Earth being spherical and not flat of infinite extent the path you take is elliptical not parabolic, not very different for long jumpers but you should keep it in mind if you are launching rockets.

At about the age of 11, I had a cliff (about 300ft) at my disposal for a range of activities, including stone throwing. One thing I learned, quickly, was that throwing stones of similar weight from different heights made little, or no, observable difference to their horizontal travel. I satisfied myself (practically) that I knew why, but paid no attention to the maths.
Now, 66 years later, here are all the relevant figures. Aint life surprisin’?