Calculation of Gravity vs Long-Jumper

Reply Thu 28 Aug, 2008 09:03 pm
A friend and I disagreed about a basic physics question, but neither of us knew how to do the math. Basically, the hypothetical was a runner who barely makes a ten foot jump on level ground compared to the same jump if the runner were jumping onto a platform five feet below him. Obviously the fall would allow the jumper to move a little further forward before landing, but we disagreed whether this distance would be substantial (specifically, a five-foot difference).

I was hoping someone here might be willing to impart the math knowledge I am lacking.

I am also curious to compare the extra distance for a jumper who can jump 15 or 20 feet on level ground.

Let me know if the problem requires more information, but any calculations would be appreciated.

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Reply Thu 28 Aug, 2008 09:35 pm
I never was any good at exterior ballistics. No calculations, but forward velocity will drop quickly, and your jumper isn't going to gain much distance at all.

Welcome back, SCoates, and where's that blob of pink bubble gum that used to be your avatar?
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Reply Thu 28 Aug, 2008 10:51 pm
If he's only jumping 10 feet then he's hardly a long jumper. The record now rests at well over 29 feet.
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Robert Gentel
Reply Thu 28 Aug, 2008 11:19 pm
I am too lazy (and don't know enough physics to do this easily) to figure this out and I don't know the variables (like velocity, takeoff angle) for a long jumper, but off the top of my head I think the simplest way to calculate the approximate extra distance can be calculated as:

How much extra distance = x*y

x = jumper's feet per second
y = seconds it takes jumper to freefall 5 feet

But if you want to do it all, with the takeoff angle and all I'd start with this:


That should tell you a bit about how to figure out the optimum takeoff angle. I'm not sure you want to get that specific but you may want to factor that in as a difference, because the long jumper can change the angle based on the very different conditions and because the take off speed of the jumper changes based on take off angle a lower takeoff angle's higher velocity might make a difference if the lower platform allows for a lower takeoff angle.

And once you have all the variables figured out, this PDF seems to walk through exactly what you are trying to do:

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Reply Thu 28 Aug, 2008 11:28 pm
Roger, I don't know where it is. The site is suddenly new and scary.

Nick, for his size, 1- feet is actually VERY impressive.

Robert, still seems too complex for me, but thanks.
Robert Gentel
Reply Thu 28 Aug, 2008 11:58 pm
Ok, then try this:


They even have a little game you can play around with and change.

But if you want my guess (without doing any calculations) I think almost 5 feet additional distance is about right.
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Nick Ashley
Reply Fri 29 Aug, 2008 11:13 am
Speculating is more fun then calculating, anyways!

Initially I agreed with Roger, thinking that they would be running out of forward momentum by the end of the jump. However, then I watched some videos on youtube, and have changed my mind. They still have alot of forward momentum when they land:

It seems like they don't get much height in their jump. Why is that? Wouldn't one go the farthest by jumping at a 45 degree angle? I would think the maximum distance anyone can jump is all their strength at a 45 degree angle, so that when they hit the ground, they have essentially no more forward momentum. My gut tells me that all the force they land with is wasted, and if they would have applied more power at an upward angle, it would have been used to propel them farther.
Robert Gentel
Reply Fri 29 Aug, 2008 12:04 pm
@Nick Ashley,
This link goes into what the optimum take off angle is.

The basic reason it's not 45 degrees is that the "optimum is 45" fact assumes equal velocity at all angles but with long jumpers the higher the angle, the lower the velocity. It says the optimum angle is around 20 degrees (and different for each jumper).
Reply Fri 29 Aug, 2008 01:02 pm
@Robert Gentel,

Let's assume 20 degrees upward angle for launch

The velocity vector can be separated into a horizontal component and a vertical component.

So vector Vtotal = vector Vvertical + vector Vhorizontal

By pythagorean theorem we get:

Vvertical = Vtotal * sin(20)

Vhorizontal = Vtotal * cos(20)

Vertical distance will net to zero, and is goverened by the equation distance = Velocity * time + 1/2 * acceleration * time^2. Acceleration is -9.8 meters/second^2. (Negative because intitial velocity is up and gravity is pulling down.)


0 = Vvertical * hangtime + 1/2 (-9.8 meters/second^2) * hangtime^2

0 = Vtotal * sin(20) * hangtime - 4.9 meters/second^2 * hangtime^2

Divide by hangtime

0 = Vtotal * sin(20) - 4.9 meters/second^2 * hangtime

Vtotal = (4.9 meters/second^2 * hangtime)/sin(20)

Next, compute horizontal component:

Distancehorizontal = Vhorizontal * hangtime

(I'm gonna cheat a little and say 10 feet = 3 meters when it really equals 3.048 meters)

3 meters = Vtotal * cos(20) * hangtime

Vtotal = 3 meters /(cos(20) * hangtime)

Substitute Vtotal above

(4.9 meters/second^2 * hangtime)/sin(20) = 3 meters /(cos(20) * hangtime)

Solve for hangtime

hangtime = squareroot(3 meters * tan(20)/4.9 meters/second^2)

Which is 0.47205817325674093410684661156673 seconds.

This means that Vhorizontal = 6.355 meters/sec

Vhorizontal = 6.355 meters/sec = Vtotal * cos(20)

Vtotal = 6.763 meters/sec

Vvertical = Vtotal * sin(20) = 2.313 meters/sec

Now, how much time will an additional 5 feet drop add to the hangtime?

Fortunately, there's a handy calculator http://hyperphysics.phy-astr.gsu.edu/Hbase/mot.html (scroll down to "motion example).

add'l hangtime = 0.36550106447852343 seconds

Additional distance covered = 2.32275926476101639765 meters = 7.62060126 feet.

Reply Fri 29 Aug, 2008 01:05 pm
Note that the results are skewed because of the short jump. For a real longjump, five add'l feet of drop would not add nearly as much.
Reply Fri 29 Aug, 2008 01:15 pm
OK, I was wrong. I'm big enough to admit it.

A nine meter jump (29.5 feet or so) has a hang time of 0.81762874020882712909950263084858 seconds.

Vhorizontal = 11.007440855982322401378209643276 meters/second
Vtotal = 11.713873890779615467089632246437 meters/second
Vvertical = 4.006380827023252932587562891158 meters/second

Add'l hangtime = 0.27911850320769976 seconds

Add'l horiztonal distance = 3.0723804158690672471377739394586 meters = 10.0799882 feet

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Reply Fri 29 Aug, 2008 01:16 pm
And for God's sake, somebody double-check my math.
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Reply Fri 29 Aug, 2008 03:26 pm
Wow, guess I was wrong.

Actually, way wrong.
Reply Fri 29 Aug, 2008 03:29 pm
If, by any rare chance, you want to do another problem, I'm curious how HIGH the platform he was jumping onto would have to be for him to LOSE five feet off of his jump, and then at that same height, how many extra feet he would gain on the way back.

Just such a situation might help me salvage a victory out of the argument.
Reply Wed 5 Nov, 2008 11:30 pm
Using feet throughout (g=-32 ft/sec/sec), I get a jump of 17.7426 feet which confirms the previous result which involved rounding.

The platform would have to be 0.9099 feet high to reduce the jump to 5 feet. Jumping from this platform would yield a distance of 12.0711 feet.
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