# Explaining the physics of the motion of falling objects

May 21, 2012 7:30 PM Subscribe

How would explain and draw the solution to this physics problem re: packages dropped from an airplane (projectile motion/falling objects)?

Why would the distance between the packages dropped from the plane increase if the package would both be under constant acceleration due to gravity, with their velocities increases at the same rate? I can't visualize this or draw it out in a way that makes sense.

Is it because the first package will have its increase in velocity sooner, causing it to travel farther over the same time interval vs the second package dropped?

*A package of supplies is dropped from a plane, and one second later a second package is dropped. Neglecting air resistance, the distance between the falling packages will*

a) be constant.

b) decrease.

d) depend on their weight.

a) be constant.

b) decrease.

**c) increase.**d) depend on their weight.

Why would the distance between the packages dropped from the plane increase if the package would both be under constant acceleration due to gravity, with their velocities increases at the same rate? I can't visualize this or draw it out in a way that makes sense.

Is it because the first package will have its increase in velocity sooner, causing it to travel farther over the same time interval vs the second package dropped?

Their velocities do increase at the same rate, but the first package as extra starting velocity of g(1s) relative to when the 2nd package is dropped. So from the 2nd package's perspective the 1st package is traveling away at g(1s) m/s.

But I had to write out the equations, then look at it and realize the obvious answer. So go with the math if you're confused.

posted by Chekhovian at 7:40 PM on May 21, 2012

But I had to write out the equations, then look at it and realize the obvious answer. So go with the math if you're confused.

posted by Chekhovian at 7:40 PM on May 21, 2012

Wow, yeah, you have the right answer, but that question throws all sorts of visualization challenges at you.

Rephrase it as: two rockets are released in a straight line away from you. Both have engines that will burn forever and give them constant acceleration. One rocket is released a second before the other. Does the distance between the two rockets...

As you noted, the first rocket travels

After that, both objects increase their velocity by another

The description makes you want to intuit what would *really* happen, which is that the two objects would both have a parabolic arc slowed by wind resistance and displaced by the movement of the airplane, all of which is totally irrelevant to what the question is trying to ask.

posted by meinvt at 7:40 PM on May 21, 2012

Rephrase it as: two rockets are released in a straight line away from you. Both have engines that will burn forever and give them constant acceleration. One rocket is released a second before the other. Does the distance between the two rockets...

As you noted, the first rocket travels

*a/2*distance and accelerates to velocity*a*in the first second.After that, both objects increase their velocity by another

*a*every second. So, the first object will always be moving away from the second at a constant velocity of*a*.The description makes you want to intuit what would *really* happen, which is that the two objects would both have a parabolic arc slowed by wind resistance and displaced by the movement of the airplane, all of which is totally irrelevant to what the question is trying to ask.

posted by meinvt at 7:40 PM on May 21, 2012

Best answer: Acceleration due to gravity is 9.8m per second per second.

The first package, at the drop, starts to fall. After 1 second, it's at 10m/s, and has traveled 10m. The second package hasn't fallen yet.

After two seconds, the first package is at (2*10m/sec=20m/sec), and has traveled (10m+20m=30m) The second is at 10m/sec, and has fallen 10m. They are now 20m apart

After three seconds, the first package is at (3*10m/sec, or 30m/sec). It has traveled (10+20+30m=60m) The second is at 20m/sec, and has traveled 30m. They are now 30m apart.

After four seconds, the first is at (4*10m/sec) 40m/sec, and has travelled 10+20+30+40m=100m. The second has travelled 60 meters (10+20+30), and they are now 40m apart.

posted by eriko at 7:43 PM on May 21, 2012

^{2}, commonly written as 9.8m/s/s, or 9.8m/s^{2}. But, for this example, I'm going to make it 10m/s^{2}, because the math is easier and I have beer.The first package, at the drop, starts to fall. After 1 second, it's at 10m/s, and has traveled 10m. The second package hasn't fallen yet.

After two seconds, the first package is at (2*10m/sec=20m/sec), and has traveled (10m+20m=30m) The second is at 10m/sec, and has fallen 10m. They are now 20m apart

After three seconds, the first package is at (3*10m/sec, or 30m/sec). It has traveled (10+20+30m=60m) The second is at 20m/sec, and has traveled 30m. They are now 30m apart.

After four seconds, the first is at (4*10m/sec) 40m/sec, and has travelled 10+20+30+40m=100m. The second has travelled 60 meters (10+20+30), and they are now 40m apart.

posted by eriko at 7:43 PM on May 21, 2012

Best answer: Gravity is 32 feet per second per second, so after 1 second the velocity of the package is 32 feet per second, and every second after that the velocity of the package increases by another 32 feet per second. The first package that was dropped has had more time to gain velocity than the 2nd package at any point in time. Because the first package is always falling faster than the 2nd package the distance between them keeps increasing.

posted by Sal and Richard at 7:44 PM on May 21, 2012

posted by Sal and Richard at 7:44 PM on May 21, 2012

Correct, the first package will be accelerating earlier than the second package, and thus be at a higher velocity at any given point in time. Thus the first package has an ever increasing head start.

Check it:

https://docs.google.com/spreadsheet/ccc?key=0AvWtOUIyk-5WdDNKQ0VTb3QzRy1Sdk1rb0dLVEsySUE

http://www.grc.nasa.gov/WWW/K-12/airplane/mofall.html

posted by wrok at 7:45 PM on May 21, 2012

Check it:

https://docs.google.com/spreadsheet/ccc?key=0AvWtOUIyk-5WdDNKQ0VTb3QzRy1Sdk1rb0dLVEsySUE

http://www.grc.nasa.gov/WWW/K-12/airplane/mofall.html

posted by wrok at 7:45 PM on May 21, 2012

Correct. The first package will always be traveling faster downwards, because it had a headstart with the acceleration, therefore the distance will always be increasing.

At all times, the packages are increasing their velocity by the same amount, so when the second package is traveling n speed, the first will be traveling n+y. (I'd say that Y should be constant, even thought the speeds are increasing)

Normally air resistance would create a terminal velocity, but that has been removed from the equation

To visualize it, perhaps plot the distance of each from the drop altitude for 1 second increments of time. One gets a head start, so every second, it's speed will be more than the other, so it's total distance traveled ahead will keep increasing.

posted by -harlequin- at 7:46 PM on May 21, 2012

At all times, the packages are increasing their velocity by the same amount, so when the second package is traveling n speed, the first will be traveling n+y. (I'd say that Y should be constant, even thought the speeds are increasing)

Normally air resistance would create a terminal velocity, but that has been removed from the equation

To visualize it, perhaps plot the distance of each from the drop altitude for 1 second increments of time. One gets a head start, so every second, it's speed will be more than the other, so it's total distance traveled ahead will keep increasing.

posted by -harlequin- at 7:46 PM on May 21, 2012

After the first package hits the ground, the distance between the packages will decrease. So, answer b) is also correct!

posted by thelonius at 8:42 PM on May 21, 2012 [3 favorites]

posted by thelonius at 8:42 PM on May 21, 2012 [3 favorites]

Eriko is conceptually correct but his math is wrong. If a package accelerates at a constant rate from a stop to 10m/s in one second, then at no point was it ever going faster than 10 m/s and at some point it was going slower than that, hence it is impossible for it to have travelled 10 meters in its first second of motion (in this case, it actually travels 5 meters).

posted by tylerkaraszewski at 8:59 PM on May 21, 2012

posted by tylerkaraszewski at 8:59 PM on May 21, 2012

Best answer: Alright, time for the Algebra Hammer:

(distance package two falls) = (1/2)g t^2

(distance package one falls) = (1/2)g (t+1)^2

see what I did there? The total time that package one has fallen at any point is the time that package two has been falling +1 second.

(distance package one falls) - (distance package two falls) = (1/2)g (t+1)^2 - (1/2)g t^2

= (1/2)g [ t^2 +2t +1 - t^2] = (1/2)g (2*t + 1)

So the distance between them is increasing linearly in time.

BOOM!

Suck that NUMBERS!

posted by Chekhovian at 9:58 PM on May 21, 2012 [4 favorites]

(distance package two falls) = (1/2)g t^2

(distance package one falls) = (1/2)g (t+1)^2

see what I did there? The total time that package one has fallen at any point is the time that package two has been falling +1 second.

(distance package one falls) - (distance package two falls) = (1/2)g (t+1)^2 - (1/2)g t^2

= (1/2)g [ t^2 +2t +1 - t^2] = (1/2)g (2*t + 1)

So the distance between them is increasing linearly in time.

BOOM!

Suck that NUMBERS!

posted by Chekhovian at 9:58 PM on May 21, 2012 [4 favorites]

And thelonius, I would have given you some points had I been marking your quiz. Even wrong cleverness always beats wrong boring. The engineering students that just put down the wrong answer with no explanation would get a big fat null.

posted by Chekhovian at 10:01 PM on May 21, 2012 [1 favorite]

posted by Chekhovian at 10:01 PM on May 21, 2012 [1 favorite]

And for extra credit:

= (1/2)g (2*t + 1) simplifies to = gt + (1/2)g (1s), differentiate that and relative velocity is just g*(1s).

From the rocket perspective suggested by

You have to be careful about the dimensions here, remembers that 1 in the time equation is really 1s, it would be better to rewrite everything starting from t + tsep or something. Otherwise you spend 5 minutes staring at g*t wondering why it doesn't have the proper dimensions [distance] when its really g*t*tsep...not that I would ever be so foolish as to do that or anything...

posted by Chekhovian at 10:50 PM on May 21, 2012

= (1/2)g (2*t + 1) simplifies to = gt + (1/2)g (1s), differentiate that and relative velocity is just g*(1s).

From the rocket perspective suggested by

**meinvt**, this just means that package 2 sees package 1 receding at a constant velocity g*(1s).You have to be careful about the dimensions here, remembers that 1 in the time equation is really 1s, it would be better to rewrite everything starting from t + tsep or something. Otherwise you spend 5 minutes staring at g*t wondering why it doesn't have the proper dimensions [distance] when its really g*t*tsep...not that I would ever be so foolish as to do that or anything...

posted by Chekhovian at 10:50 PM on May 21, 2012

You can see this in action in the printing water fountains which are popular in Japan, such as this one on YouTube--note how the shapes sort of "stretch" as they fall. (Also there is a better video with a longer drop which shows this more clearly out there somewhere, but I can't find it any more.)

posted by anaelith at 12:49 AM on May 22, 2012

posted by anaelith at 12:49 AM on May 22, 2012

Ignore the first package's acceleration altogether and remove gravity from the situation. It's entirely legitimate to transform a uniform gravitational force away.

Change your frame of reference to the first falling package. Anything that is stationary in its frame and remains in free-fall with it remains stationary in its frame.

From this perspective the second package is accelerated away for the first second then released, and then is going to continue moving away at the velocity it gained in the first package's frame.

posted by edd at 3:47 AM on May 22, 2012

Change your frame of reference to the first falling package. Anything that is stationary in its frame and remains in free-fall with it remains stationary in its frame.

From this perspective the second package is accelerated away for the first second then released, and then is going to continue moving away at the velocity it gained in the first package's frame.

posted by edd at 3:47 AM on May 22, 2012

Forget the equations for a second.

It's like if we played a game where you stood on a street, and every hour I would tell you a number of blocks to walk, then you walk that many blocks (always in the same direction) and wait. That's your change of position.

In this case, each hour I tell you to walk 1 block, then 2 blocks, then 3 blocks, then 4 blocks, etc. The change in how many blocks you walk—the rate of increase of your velocity—is constant, like gravity.

Now, say you're being chased by someone, who starts off an hour after you. I tell them 1, 2, 3, 4, ... blocks too. Just like you, except one hour behind in the program.

So whenever I tell you to move N blocks, I'm telling the person chasing you to move (N-1) blocks. You gain a block of distance each hour.

posted by fleacircus at 3:48 AM on May 22, 2012 [2 favorites]

It's like if we played a game where you stood on a street, and every hour I would tell you a number of blocks to walk, then you walk that many blocks (always in the same direction) and wait. That's your change of position.

In this case, each hour I tell you to walk 1 block, then 2 blocks, then 3 blocks, then 4 blocks, etc. The change in how many blocks you walk—the rate of increase of your velocity—is constant, like gravity.

Now, say you're being chased by someone, who starts off an hour after you. I tell them 1, 2, 3, 4, ... blocks too. Just like you, except one hour behind in the program.

So whenever I tell you to move N blocks, I'm telling the person chasing you to move (N-1) blocks. You gain a block of distance each hour.

posted by fleacircus at 3:48 AM on May 22, 2012 [2 favorites]

The equation of motion for both packages is the same (v=v0+at). The only thing different is the first package has a greater initial velocity. So at any given time the first package will have a greater velocity, hence the distance between them must increase.

posted by azathoth at 7:46 AM on May 22, 2012

posted by azathoth at 7:46 AM on May 22, 2012

One other point, a lot of people approach these issues think that they need to understand everything intuitively and automatically, without relying on any math. Most people can't do this very well, and to demand it of yourself is setting a harsh and unrealistic standard.

No one serious does this. What usually happens is you solve the math, then realize the answer was "obvious" in retrospect, if only your mind had been the right place. Usually it isn't, but thats why we have math. Actually it sort of becomes an iterated feedback loop, math, physical intuition, math...etc

But don't fall into the "engineer's trap" of putting numbers in at the first step and relying solely on your calculator for all your conclusions. Don't be that sort of monkey.

posted by Chekhovian at 4:16 PM on May 22, 2012

No one serious does this. What usually happens is you solve the math, then realize the answer was "obvious" in retrospect, if only your mind had been the right place. Usually it isn't, but thats why we have math. Actually it sort of becomes an iterated feedback loop, math, physical intuition, math...etc

But don't fall into the "engineer's trap" of putting numbers in at the first step and relying solely on your calculator for all your conclusions. Don't be that sort of monkey.

posted by Chekhovian at 4:16 PM on May 22, 2012

This thread is closed to new comments.

It's easy to confuse constant acceleration with constant speed.

posted by RobotHero at 7:39 PM on May 21, 2012