I was wondering if anyone can tell me the formula to figure out how long one will long jump based on speed and takeoff angle? Maybe it is not that simple. Just wondering what it is. Thanks.
Parabolic Equation/Formula


For practical application in long jumping, it’s not that simple. If you’re talking physics theory and projectiles, on the other hand, then it’s relatively easy to figure out.

I was wondering if anyone can tell me the formula to figure out how long one will long jump based on speed and takeoff angle? Maybe it is not that simple. Just wondering what it is. Thanks.
I had one for speed & peak height but I don’t know where my documentation is. Lets just say that if you can takeoff at 10m/s and put together any sort of takeoff (talking ~12″ vertical), you are generally over 7m (unless you are like 4′ tall).

Actually I just worked a problem using Nick Newman’s numbers and it seems very accurate. You have to take into consideration that the optimal takeoff angle for a long jumper is ~21 degrees.
I looked at Nick’s log and his best long jump is 7.54m and I believe he said somewhere he runs a 10m fly in ~.95 seconds. I’m going to use Nick’s 10m fly velocity as his take off velocity, or Vo.
First solve for time in the air: (I’m using upper case for variables and lower case as subscript)
Vy = Voy*sin() + Ay*Tup
0 = 10.526*sin(21°) + (9.8)*Tup
Tup = ~.385 seconds
Ttot = ~.770 secondsNow solve for distance / range:
X = Vx*cos()*Ttot
X = 10.526*cos(21°)*.770
X = 7.565m7.565m is pretty close to 7.54m, and if you were to take air resistance into account it would probably be even more accurate.

Using the same principles above I derived a function of distance with respect to 10m fly time:
Distance — Time
8.952m — .85 seconds
8.752m — .86 seconds
8.558m — .87 seconds
8.371m — .88 seconds
8.190m — .89 seconds
8.015m — .90 seconds
7.845m — .91 seconds
7.680m — .92 seconds
7.521m — .93 seconds
7.366m — .94 secondsI used these functions to generate the numbers above:
Time = (2(0Velocity*sin(21.1°)))/(9.80665 – Air Resistance)
Distance = Velocity*cos(21.1°)*Time
Velocity = 10 / ((10m fly time) + .02)
Air Resistance = (.5(Velocity*sin(21.1°))^2) / 79.3786647 
Looks somewhat logical but remember no one is taking off at their PR fly 10m velocity.
I know if you look at the function for velocity I’m using their fly 10m PR + .02 seconds which is a guess, but produced good numbers.

Actually I just worked a problem using Nick Newman’s numbers and it seems very accurate. You have to take into consideration that the optimal takeoff angle for a long jumper is ~21 degrees.
I looked at Nick’s log and his best long jump is 7.54m and I believe he said somewhere he runs a 10m fly in ~.95 seconds. I’m going to use Nick’s 10m fly velocity as his take off velocity, or Vo.
First solve for time in the air: (I’m using upper case for variables and lower case as subscript)
Vy = Voy*sin() + Ay*Tup
0 = 10.526*sin(21°) + (9.8)*Tup
Tup = ~.385 seconds
Ttot = ~.770 secondsNow solve for distance / range:
X = Vx*cos()*Ttot
X = 10.526*cos(21°)*.770
X = 7.565m7.565m is pretty close to 7.54m, and if you were to take air resistance into account it would probably be even more accurate.
While largely true an addon needs to be made on to these comments. There are three factors which are neglected in this equation:
*The height of the COM at takeoff and the COM at landing is quite different (about 1m). I’d guess this would add about 0.3m or so.
*The location of the COM at takeoff is actually in front of the point of measurement. Parabolic flight equations begin when the body or object becomes airborne regardless of some relatively irrelevant point like the takeoff board. I’d guess this could add up to 0.3m on to the mathematically projected distance.
*The point one makes contact with the sand can actually be in front of the COM which like the above point isn’t accounted for in projectile motion equations. I’d guess this can add up to 0.3m.
ELITETRACK Founder

Did your equation take into account hip height at takeoff or does your equation base the projections off projectile distance from even ground.
I use an equation to project long jump distance except I based the equation with respect to hip height at takeoff as well. So it has speed, take off angle and hip height at ground contact to produce x result. This has worked well for both my male and female long jumpers to show them if hip height, speed or take off angle are altered it would produce x result in comparison to what they are currently doing.
Mike do you know if at the Trials they had any markers down near the takeoff for the long jump like they did for the pole vault in the finals. If there was markers done using dart fish we could attempt to get some hard data to back any equations developed from professionals

My equations did not take into account COM, as Mike pointed out. I’ll be able to spit out some more accurate functions when I’m finished with this semester of AP Physics :P.

yeah mike is right….
there is no way someone should jump 7.56m running 10.5m/s…
i have somewhere a chart where this has been figured out. I am not sure where but i will look.
But 10.5 m/s with a take off angle of 21 degrees will be over 8 metres.
and just to add, when i jumped 7.54m i wasnt running anywhere near 10.5 m/s . . .last year i was closer to that speed, however fouls and bad landings created many problems…

There is really only one projectile motion equation concerning distance. We can’t really make something up and expect it to be better. My point in the previous post was just to point out that when you use the projectile motion equation to determine distance you need to recognize the 3 points noted above. Also, as Mort pointed out, the 10m fly time velocity isn’t even what you’d want to put in to the equation. If you wanted to do it accurately, you’d get the instantaneous velocity at the moment of takeoff (which would be in front of the board).
The horizontal jumps at the trials had markers and data collected over the last 20m at 1000 fps (yes 1000). I am not sure what will come of this video though.
Unfortunately dartfish wouldn’t be able to give you anything accurate. Despite what the dartfish reps try to sell, it’s practically worthless for determining angles, distances and velocities.
ELITETRACK Founder

Yeah Nick was probably taking off around 10.2 m/s (not fly but takeoff velocity) with several jumps (all fouls) that were takeoff point to touchdown point 7.75 or greater.
ELITETRACK Founder
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