Dot Physics

Hancock throws a boy. Not nice.

I finally saw the movie Hancock. Yes, I know it has been out for a long time but I don’t get out much. You know me, I can’t leave something like this well enough alone. It’s not my fault, I was born this way. It shouldn’t spoil the movie too much if I tell you this one scene (you have probably already seen it anyway).

i-ff47848a75255e93595c1fdbf563bc64-hancock_pic_1.jpg

Basically, Hancock gets upset with this boy and throws him in the air to scare him or something. In case you did not time it, the kid was in the air for 23 seconds. I claim that in order for Hancock to throw a person in the air for this long, the acceleration during throw would be deadly.

For the first pass, what if there were no air resistance (clearly, there is). In this case, I can determine the initial velocity of the boy and from that his acceleration during ‘the throw’. If the time the boy is in the air is t, then I can use the definition of acceleration:

i-103385c5497c471358d97448cfea2153-la_te_xi_t_1_12.jpg

If the boy is thrown up and falls with a constant acceleration (g) then his final velocity will be the opposite of what his initial velocity was. From this, I can solve for the initial velocity:

i-5d60157f43c5110ea6bd502634ba95b6-la_te_xi_t_1_13.jpg

For 23 seconds, this gives an initial velocity of 113 m/s (or 250 mph). Clearly this is fast enough that air resistance will come into play. But already, you can see if this boy is accelerated from 0 m/s to 113 m/s in the distance of about 2 meters (or less) then there is going to be trouble.

I think I have already shown my point, but that is not enough. I need to take it to the next (but not the final) level. If I include air resistance, how fast would Hancock have to throw this kid so that he is in the air for 23 seconds. Assumptions:

  • I am going to assume the boy has the same terminal velocity as a grown man. This will allow me to use my sky diver falling model (from ‘can an iphone tell if your parachute didn’t open’) without much modifications. I would imagine that a smaller boy would fall about the same as a grown man because he would have both smaller surface area and mass (although these don’t change the same with scaling).
  • Position. In the clip, the boy seems to come down in sky diving position, but he looks like he is thrown up in “feet down” position. This could make a difference, but I am going to model this as though the boy had the same position during the whole flight.
  • Assume that the air density is constant. Of course, it isn’t – but should be close enough to constant for this. Also, this can easily be changed later.
  • Finally, I will assume that the gravitational field is constant.

Ok, now on for the calculation. The basic plan is to:

  • Calculate the force on the boy while in the air. This will be the gravitational force plus the air resistance. In one dimension, I need to make sure that the air resistance force is in the opposite direction as the motion.
  • Calculate the acceleration. (a = Fnet/m)
  • Update the velocity. (v = v + a*dt)
  • Update the position. (y = y +v*dt)
  • Update the time.
  • repeat
  • Plot stuff

That is the basic idea. If you want help with numerical calculations, check out my previous introduction. Anyway, here is a plot of the boy being thrown up with an initial velocity of 113 m/s (blue line). I have also plotted (for comparison) an object with no air resistance (green line).

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Both lines represent an object thrown up with the same speed. You can see that the air resistance case does not go as high (because of the air resistance). And even though it is going much slower on the way down, it still is not in the air as long as the no-air resistance case.

Next question: how fast WOULD he have to be thrown to be in the air for 23 seconds? To answer this question, I am just going to put another step in to the program. I will run it at 110 m/s, then 115 m/s then 120 m/s and so on. For each “run” I will have the program record the time. Simple, no?

Here is a plot of the time of flight for a “sky diver” with an initial upward velocity of 5 m/s to 1000 m/s.

i-83acdaf0fdc1f3b951de672ff7a928a3-figure_1_1.jpg

From this graph, it seems the initial velocity for a thrown skydiver would need to be around 400 m/s in order for him (or her) to be in the air for around 23 seconds. You can see also, that this curve starts to “level off” in that to increase the flight time (or hang time if you like basketball) take a greater and greater initial velocity. Let me go ahead and re run this up to an initial velocity of 5000 m/s, you know ….just because.

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By increasing the initial velocity from 1000 m/s to 5000 m/s, the flight time only increases by about 10 seconds. This is because at such high speeds, there is a tremendous air resistance force that quickly slows the skydiver down. Oh, one more thing on this part. Recall the first part above where I showed the time of flight without air resistance. Without air resistance, this graph of time of flight would be a straight line (ignoring changes in gravitational field).

Now I am ready for the second part. Let me use 400 m/s as the initial velocity of the kid to be in the air for 23 seconds. What would his acceleration be during the “throw” from Hancock? Here, I am in the situation where I am just interested in acceleration and distance and not time. Usually, I would automatically think of the work-energy theorem. However, by manipulating the kinematic equations, I can get an expression without time.

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For the boy, his initial velocity is 0 m/s. Solving for the acceleration, I get:

i-c56cc5ec00efe5c744e4a6dde417e5a3-la_te_xi_t_1_21.jpg

Plug in what values you think are reasonable. I am going to use a final velocity (final for the throw is initial for the in the air part) of 400 m/s and a distance of 1.5 meters (which I think is quite generous). This gives an acceleration of over 50,000 m/s2. If you like this in terms of “g’s” then this is like 5000 g’s. Danger.

This table of NASA g-force tolerance data used to be on wikipedia’s page, don’t know why they took it off, but here it is:

i-4107cd4992b2d9e84b6ca0d7693546b6-iron_man_physics_and_g_tolerances_dot_physics.jpg

If the boy is being thrown while facing down, that would be “eyeballs out”. Notice that nowhere on the table is there a tolerance anywhere near 5000 gs in any position for any time. The result would be a dead bully.

Comments

  1. #1 Alex
    June 30, 2009

    Another fun analysis but I think you have a typo. In the case of no air resistance the acceleration during the throw would be about 4,000 m/s^2. With air resistance (using your numbers) you get an acceleration of about 50,000 m/s^2.

  2. #2 Rhett
    June 30, 2009

    @Alex,

    Thanks for pointing out the error. I fixed it.

  3. #3 Uncle Al
    June 30, 2009

    The young will never believe Enviro-whinerism, God, or the Federal budget if they pause to do the algebra. Homeland Severity is not recursive terrorism, dialing 911 saves you,

    http://www.bestviral.com/i/images/138.jpg

    Supreme Court nominee and self-declared “wise Latina woman with the richness of her experiences” Sonia Sotomayor demands that evaluation containing literacy or numeracy requirements actively discriminates against minorities (e.g., firefighters, voting; conceivably high school diplomas). Would you have us believe scarce survivor of an ancient race of Angels of Colour Hancock goes around swilling malt liquor and abusing male children? Who died and made him… Oh, yeah. Never mind.

  4. #4 Jeff
    July 3, 2009

    I think there may be an error in your initial analysis. You say that the boy spent 23 seconds in the air. This means he spent 11.5 seconds up and 11.5 seconds down. If we use the basic kinematic equations for a falling body we can model only the second half of the flight to get initial velocity (think of it as dropping the boy from the max height, not throwing him up, at max he will have zero velocity so its the same): d=.5at^2 and v=d/t; we can do a similar analysis. Using the above equations we get vt=.5at^2, this means that v=.5at (Just as you got). It is my contention that you should use the time of 11.5 (NOT 23). You would therefore get an initial velocity half of what you got, or 56.5 m/s.

  5. #5 Rhett
    July 4, 2009

    @Jeff,

    You fell victim to one of the classic blunders! The most famous is “Never get involved in a land war in Asia,” but only slightly less well known is this:

    v-average = d/t, not v=d/t. The velocity is not constant in this case, so you analysis is flawed.

  6. #6 ryan
    April 17, 2011

    But you aren’t taking into account that hancock might have some sort of power like some sot of telekinesis that can create a protective aura. Since he’s an “angel” he might have a myriad of unknown powers

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