Projectiles: Newsletter Challenge (04/01/08)

v+9.81t²

Regards JD

Consider the idea that the projectiles are bullets. Unless the first bullet is traveling very slowly, I do not believe you can discount air resistance. One of the major limiting factors in the speed of a bullet, is the air resistance. It is entirely possible that both bullets had the same initial velocity but the first one has much more air resistance than the second, slowing it down quicker.

Of course they could be launching water melons with a giant sling shot. But that could get messy.

As I think Physicist? mentioned in a previous challenge it is assumed that the question has all of the information required to solve it otherwise there can be no unique answer. In this case if friction was to be included we would need additional info such as the aerodynamic profile/drag of each projectile, the mass of each projectile, the initial velocity of projectile A, and perhaps the air pressure. I suspect even then that someone would then cite a situation on another planet with a different atmosphere. I guess thats part of the lateral thought that comes with these questions.

I suppose you could say that, but you would think the question would say to ignore friction losses. And as for other planets, we certainly have had arguements about that.

Hi BobD

You were faster! Just for the record I post my result which is of slightly different form:

v₂ = v + (gt)² / 2(v-gt)

It's equivalent to yours, only that it requires slightly less computations (I'm a software guy; can't help it!)

Anyway, it's a shame that the challenges have become so easy as to be solved after a couple of posts or so...

2.4.08

BobD used to be my best friend in school. Often I used to teach him maths and physics!

Dont understand the application of the term v₂((v/g)-t) for the distance B travels at constant velocity v₂ - if gravity is effective immdiatly upon discharge There is no distance B travels at constant velocity. Also if applicable why not apply it to A as well??

Without it the solution reduces to

v₂²/2g = 1/2.g.((v/g)-t)²

Which ends up as

v₂ = v - tg

and realizing that g is negative vector in the system becomes

v₂ = v + tg for the numerical solution.

The equation for the distance s, travelled by a projectile travelling upwards against gravity g, with initial velocity u in time t is :-

s = ut - ½gt²

If gravity g was zero as in outer space then this reduces to s = ut

or

Distance = Velocity X time

BobD is indeed correct

The formula for distance travelled by an object with an initial velocity and subject to constant acceleration (assuming velocity is low enough that we do not have to take relativistic effects into account) is D = ut+1/2.a.t² where u = initial velocity (vector), a = acceleration (vector) and t = time over which distance travelled is measured. Where there is no initial velocity the distance travelled is just the result of the acceleration and when there is no acceleration distance is just the result of constant velocity u. In this case the projectile has an initial velocity (otherwise it would not leave the ground) which needs to be included in the calculation.

As to why I didn't seem to use this formula when calculating distance for A. This is because for T₁ = time for velocity to change from initial velocity v to 0, vT₁ = g.T₁², therefore D = g.T₁²-1/2.g.T₁² = ½.g.T₁² Note that this is not true for other values of T₁.

This can be proved by using the standard formula as follows:

D = vT₁ – ½.g.T₁² and substituting T1 = v/g (because we know that over this time the velocity changes from v to 0 as a result of constant acceleration g) gives

D = v²/g – ½.g.v²/g² = 1/2.v²/g = ½.g.T₁²

Your solution of v+tg can be shown to be incorrect because when t approaches T₁, ie, the closer B is launched to when A reaches its max height the faster its velocity needs to be to reach that height in the remaining time and approaches infinity as t approaches T₁. Your solution has a value of 2v when t = T₁.

I think your solution is incorrect. You appear to have made the fatal mistake of labelling the initial velocity of projectile B as V2 instead of VB. You have then gone on to confuse this with the inital velocity of A squared also appearing on the computer as V2. This mistake has gone on to confuse your subsequent calculation.

from ORDE

Try following it through. If you can't cope with subscripts, copy it and reletter it as VB. You will find there was no mistake (though you may want to have some extra stages).

Fyz

I think your solution is incorrect. You appear to have made the fatal mistake of labelling the initial velocity of projectile B as V2 instead of VB. You have then gone on to confuse this with the inital velocity of A squared also appearing on the computer as V2. This mistake has gone on to confuse your subsequent calculation.

I repeat: try following it through. If you can't cope with subscripts, copy it and reletter it as VB. You will find the result was correct (though you may want to add some extra stages in the derivation).

Fyz

I agree with BobD on the air resistance effect on the projectiles, anyway, the problem does not specify any of this and to even think about it means that we are not reading the challenge properly. We must not forget that this is an all theoretic problem.

Regardless, the analysis by BobD is preatty neat, but I found his final result to complicated.

Following a similar analysis, here is my own analysis of the problem:

projectile A:

Instant velocity:

v' = v - g.t_a

where:

t_a: time used by projectile A to reach its highest point.

as the point to analyze is the highest point it reaches, the instant velocity there is 0

so:

v'= v - g.t_a = 0; ==> ta = v/g

this time used by projectile A to reach its highest point is our baseline for the following analysis.

Projectile B:

Instant velocity:

v" = v_0b - g.(t_a - t)

where:

v0b : initial velocity of projectile B (our target)

t: delay time to launch B after A was launched.

Now, if we make this expresison equal to 0, we are going to match the maximum height reached by projectile A.

v" = v_0b - g.(t_a - t) = 0; ==> t_a - t = v_0b/g; ==> t_a = v_0b/g - t

as this time is our original value calculated for projectile A, we can write:

v/g = v_0b/g - t

multiplying both sides by g, we arrive to:

v = v_0b - g.t

and from here:

v_0b = v + g.t

Now, this value is giving us the actual velocity to reach projectile A highest point with projectile B after a launch delay time of "t" seconds. But that is not the answer to the challenge.

I have notice that even the answers that match the last expression, left the problem at this point.

Now, the challenge says that projectile B "passes" projectile A at this point, so if we give projectile B a initial velocity of v + g.t, we would only achieve to reach the same highest point than projectile A.

Therefore, the right answer to the challenge, regardless the complexity of the formula, must include the sign greater than to actually answer the challenge, as follows:

v_0b > v + g.t

And that should be the answer.

Thanks, and keep the good job.

Regardless of any other incomprehensions: if projectile B is launched after projectile A, and is to reach the maximum height of projectile A at the same time as projectile A does, projectile B must still be going upwards as it reaches the maximum height. So no inequality is needed.

BobD and his followers were precisely correct (on the basis that we can neglect air resistance, Coriolis and related effects, non-uniform gravitation, special & general relativity, and doubtless a number of other modifying factors).

Yes, I agree, I was just re-examining my analysis because it did not seem right.

this looks like a science homework question, I thought admin was going to ban them?

1st year Dip. Mech Eng subject.

I'm surprised that the questions are getting easier.

I see you mentioned questions getting easier but yet you have no answer. What gives?

Ha! In Romania I was solving this type of "challenges" in grade 9. Part of the curriculum.

I believe we can infer a few things from the wording of the problem.

Projectile B passes Projectile A. Their paths must be close, but they are probably not identical, or else they might collide.

Because their paths are close, they probably encounter the same air friction, so we can safely ignore it.

Because B passes A, B must still be ascending when A stops ascending.

What we are not told is whether or not the projectiles were launched from the same elevation. If they were not, that makes this a trick question and introduces a factor that either makes the initial velocity of B equal to A, or much greater. If the initial velocity of B is much greater, then it will be indeterminable. If the initial velocity of both is the same, then why ask the question, except to confirm the fact?
I raise this issue to dismmiss it and to establish that the launch conditions for the projectiles are nearly identical. We are told that the launch times and initial velocities are different.

In short, don't forget to use Occam's razor. (Or is that Ockam's rasor?) (If Occam were alive today, would he use a RAZR?)

Because their velocities are different, air friction would not be the same. I think we must either ignore viscous drag and take the BobD et al solution, or we have to allow for the different masses of the two bodies, their Reynold numbers, and the viscocity and density of the air. I doubt that has an analytic solution (yawn)

Let: v1=initial velocity projectile A

v2=initial velocity projectile B

t0=delay between two launches

g =gravity

Velocity of A is v(t)=v1-g*t and it will be = 0 (projectile at highest point) for t=v1/g; the height reached by A is:

h1=∫(v1-g*t) with t=0 -> v1/g

Developing the integral we have:

h1=v1*v1/g-g*v1*v1/(2*g*g)=V1*v1/(2*g)

Velocity of B is v(t)=v2-g*t and the height will be:

h2=∫(v2-g*t) with t=t0 -> v1/g

Developing the integral we have:

h2=v1*v2/g-v2*t0-v1*v1/(2*g)+g*t0*t0/2

But h2 must be = h1. So:

V1*v1/(2*g)=v1*v2/g-v2*t0-v1*v1/(2*g)+g*t0*t0/2

and solving respect v2 we obtain:

v2=v1+g*t0

You might want to compare that with BObD's answer at #2. Your answers appear dissimilar.

As a first step, perhaps

h2=∫(v2-g*t) with t=t0 -> v1/g

should read:

h2=∫(v2-g*(t-t0)) with t=t0 -> v1/g

?

Let's test your solution :-

If t0 = 0 then v2 = v1 which is indeed the case.

However if the time delay t0 = v1/g then the velocity v2 would have to be infinite in order for projectile B to catch up with projectile A at the top of projectile A's path.

Now plugging in t0 = v1/g into your solution yields:-

v2 = v1 +g*(v1/g) = 2*v1 (This is not infinite !!!)

your solution is therefore not correct.

Try this with BobD's solution in post#2 and you will find that his v2 will equal infinity.

Though the problem has been solved satisfactorily at the beginning itself, some insight into it might be obtained by drawing a graph of height versus time.

With all the usual caveats about ignoring air friction, assuming constant g, ignoring coriolis effects, etc., we are left with a problem of motion in one dimension under constant acceleration.

There are two bodies involved. The first, 'A' has a starting velocity v_A at time t=0, which is the slope of the parabola at the origin. Its velocity reverses after a time t_R=v_A/g, when it has attained a height of h_A=v_A²/2g.

Body 'B' starts at time t=t_B (where t_B<t_R) with some velocity v_B such that during the time t_R-t_B it reaches the same height h_A. The plot for the second body is clearly a second identical but displaced parabola which intersects the first at its vertex.

If this situation is sketched out even very roughly, one can contemplate the implications of varying the parameters, such as negative values of t_B, or t_B approaching t_R. Appropriate little bits of algebra or calculus or co-ordinate geometry should provide the answer to the challenge question and its derivatives, but I better leave the graphics and expert treatment to the wizards of CR4 in case anyone feels so inclined.

=TeeSquare=

If projectile B starts first, it will surely be on its way back down when it passes projectile A.

TeeSquare,

Is this the graph you had in mind?

Regards,

MPM

Thanks MPM, for the graphic which clearly expresses what I tried conveying in words. The graph for projectile B can be slid around without rotation, such that it always passes through the vertex of A, and we will get a visual idea of the relative magnitudes (or actual, if numerical scales are included). As FYZ pointed out in post 47, if 't' is negative then 'B' will meet 'A' on its way down.

Some people like me find this kind of visual representation quite illuminating, though I am often too lazy and try to do it in my head if possible. I didn't attempt to make the figure for CR4 myself, as my efforts in the Funny Reflections post had taken a lot of time -- I'm rather slow and hopeless at negotiating with computers and internet.

Thanks to Kris for the thoughts in post 57. I guess my comment was an open-ended one, with little chance of any consensus emerging. I was just expressing my concern regarding our general carelessness and its implications.

Regards, =TeeSquare=

"...apparently honest practitioners of science or engineering may actually be dangerous to society..."

Entirely possible. Here we've had (over decades, mind you!) bridges collapse, parking garages implode, aerial walkways inside buildings come crashing down, dams burst, and every other kind of disaster you can contemplate. Some of these were defects in construction, and some due to lack of maintenance, but some were due to poor design. Anyone who hasn't someone to check their equations is like an author without an editor, which as we all know is just not done.

You say "check the equations".

This is probably a matter of terminology, but I've not found that checking (in the sense of reading through and looking for error) is very reliable. A quick read and replication is better; however, where possible I prefer to check the performance in an independent way (e.g. get your strongest rival to check the performance using a different set of numerical tools). When all that works, it comes down to having an adequate requirements specification...

Terminology indeed! The point is, anyone who relies solely on their own judgement should get what's coming to them when it proves deficient. Have it checked! Whatever it is, by whatever means necessary, if lives or livelihoods could be at stake, check the work!!! Or pay the piper...

"By whatever means necessary". We are in agreement.

But perhaps I should have been blunt. Someone (just) to check the equations" (your original wording, my addition) could be woefully inadequate.

For Projectile A

v₀ = v

s_A = vt_A – ½ at_A ²

v_A = v – at

Set v_A = 0 and solve for t_A since at the highest point the velocity becomes 0 for the projectile to change direction

a = g since only gravity is action on the projectile

t_A = v/g

For B

s_B = v_0Bt – ½gt_B²

v_B = v_0B – gt_B

t_B = t_A + t since projectile B was launched at time t after t_A

But t_A = v/g so substitute v/g for t_A in above equation for t_B

t_B = v/g + t

substitute t_B in equation for v_B and solve for v_0B

v_B = v_0B – [v + gt]

v_B must be > 0 since it passes projectile A at its highest point so

v_0B > v + gt

t_B = t_A + t since projectile B was launched at time t after t_A

Actually it is t_B = t_A - t (that is, t_B = 0 when t_A = t)

Otherwise you are in the right track. You have to take also into account that s_A=s_B and then you will arrive to the same result that BobD has given. The detail you seem to miss is that the projectile B does not just win projectile A in height (in which case the inequality you provide would be enough) , it surpasses it at the exact moment projectile A reverts its motion and heads down. Therefore the solution has to be exact.

Answer:

Initial velocity of projectile B = v + g^2*t^2/ 2(v-gt)

This problem would have been more challenging if B was launched at a time, t, prior to A.

Not really (in fact, not at all)
Time and height when B passes A are unchanged (= v_A/g and v_A²/(2g) respectively).

And as before the position of B is governed by
h_B = (v_B-g(t-t_B)/2).(t-t_B)
where t_B is the launch time of projectile B.

The solution** is unchanged - the only difference is that projectile B will be coming down when it passes projectile A, and you need to set t_B negative.
**It was first given by BobD in #2, and reformulated by tkot in #11.
v_B = v_A + (g.t_B)² / (2(v_A-g.t_B))

The only thing I can think of that was not already mentioned elsewhere** is the implication of setting t_B after projectile A has reached the top of its trajectory. Clearly, projectile B will be on its way down at time t_B. For v_A/g<t_B<2.V_A/g projectile B will be on its way down when it passes projectile A - but for t_B>2.v_A/g projectile B will still be on the way up when it passes projectile A.
**But indicated by sliding Maths_Physics_Maniac's graphs shown in #56

" For v_A/g<t_B<2.V_A/g projectile B will be on its way down when it passes projectile A - but for t_B>2.v_A/g projectile B will still be on the way up when it passes projectile A. "
Hello Fyz,

I've gotten a bit puzzled by your concluding statements about t_B being greater than v_A/g. I think that zone will produce only imaginary solutions, though someone may be able to offer interesting explanations of what they represent. If t_B = v_A/g , v_B will have to be infinite to meet the conditions of the problem. I did consider elaborating on these possibilities in my original post 43, but I suspect that my verbosity does not go down too well with the CR4 crowd.

=TeeSquare=

As I mentioned already in #74, Maths_Physics_Maniac's graphs shown in #56 is a great aid in understanding the problem. We have two downward facing bells interjecting at the highest point of the first one.

First, let me for a moment note that the parabolas interject at only one point. This can be shown if one takes the difference of the formulas (see them in #74). What you get is the distance between the two projectiles, and the interesting point is that this difference is a linear equation of time. Therefore, it can be 0 only in one point (i.e. the projectiles will meet at one point only.) Consequently, the second "bell" will always have one of its "feet" inside the first bell and the other outside. More specifically, if the second bell "starts" at the left of the first one (that is t_B<0) then it "ends" between v/g and 2v/g inside the first bell. If t_B>0 then the second bell "starts" between 0 and v/g and "finishes" somewhere on the right of 2v/g. All these second "bells" are solutions to the problem. There will be two "feet", i.e. B is at its launching position at two different times. These times are t_B and t_B+2v_B/g. In the leftmost point, velocity is positive (upwards) while in the rightmost it is negative and with the same measure as the first one.

My point now, is that you can imagine a projectile traveling back in time! If this ever were possible, then you could allow for launching time greater than v/g! This motion would produce the same curve as if another projectile was launched 2v_B/g time units before! The mathematical formulas we used have no knowledge whatsoever of our limitations regarding time, i.e. that it has to flow in one direction only. So in strict mathematical sense, all values of t_B provide solutions to the problem.

Another way one could phrase the problem, in order to allow all these solutions have physical meaning, is to just say that projectile B happens to be at it's launching point at time t_B after launching A. This might mean either it gets launched at t_B or hits the ground at t_B. In this case, BobD's formula would be valid, and provide a positive velocity in the first case and a negative in the latter.

That was a beautiful explanation of what happens when t_B exceeds v_B/g. The fact that a mathematical equation does not 'know' whether time travels forward or backward is an interesting revelation. There must be many situations where equations can be interpreted in revealing ways if only one has the requisite imagination.

As a mechanical engineer, my thinking is mostly limited to Newtonian mechanics and basic principles from nineteenth century physics. Elementary mathematics has of course been a useful tool. But people like me can get only some distant glimpses of the beauty of mathematics, though we may try to do a bit of wider general reading.

I had one enlightening experience when I was working on a geometrical problem regarding the meshing of gear teeth with non-conjugate profiles. For a number of parameters I derived fairly cumbersome equations full of trigonometric powers and roots, and an iterative procedure provided numerical solutions which were quite acceptable.

Some sixth sense told me that I should be able to improve on the procedure, and I got down to systematically simplifying the expressions using the known trigonometric identities, in a somewhat intuitive way. To my delight, all the equations could be represented in a very compact and elegant form. Moreover they had closed-form solutions, except for one obvious case which required a few steps of iteration.

I was disappointed that I could not share the details of my discovery with anyone who would be interested. My late respected professor was too busy at the time to go into details, but was pleased with my results. My thesis must have been added to the growing mountains of academic output which remain unread, after serving its purpose of fetching me a post-graduate degree. I don't care much for these degrees though, as over the years I have come to respect people for their knowledge or integrity, rather than any paper qualifications or high-sounding job designations.

Though some have condemned this challenge question as being too trivial, I find it has thrown up interesting on-topic insights.

Regards, =TeeSquare=

You are right that the solutions after v_A/g do not correspond to projectile B being "launched" at the later time T.

What they correspond to is the projectile arriving at the ground after it has passed projectile A - obviously this is a real solution for the mechanical situation; but it is not an answer to the question.

Equally, any "solution" exceeding the speed of light is void.

Maths_Physics_Maniac's graphs shown in #56 is indeed a great way to understand the problem. We have two parabolas crossing at one particular point which is the (v/g, v²/2g), i.e. the maximum point of the first parabola. The two parabolas are given by the formulas:

v_At - 1/2gt² and v_B(t-t_B) - 1/2g(t-t_B)² where t_B is projectile's B launch time (= the 't' in the original question's text)

Now - and this is the interesting bit - t_B can as well be negative! That is, projectile B may be launched before A. It is already mentioned by Physicist, I just rephrase it to draw your attention to the figure: In any case, the second parabola would own the point (v_A/g, v_A²/2g), so by substituting we get:

v_A²/2g = v_B(v_A/g - t_B) - 1/2g(v_A/g - t_B)² => ... => v_B = ... BobD's answer.

You can if you like feed t_B<0 in the formula and still get a valid answer.

The initial velocity of projectile B = v^2/2gt-1/2gt^2

The projectile A reaches highest point H= v^2/2g

This H should be = distance by the projectile B = ut -1/2gt^2 (u= initl. vel. of B)

Thus v^2/2g=ut-1/2gt^2

u= v^2/2g-1/2gt^2

Projectile A Governing Eqns:

H= vT-1/2gT^2 (H= Max height; T= time of travel)

0=v-gT Thus T=v/g (g= Accrn due to gravity)

0=v^2-2gH Thus H=v^2/2g

Projectile B Governing Eqns:

H= u (T-t)-1/2g(T-t)^2 (u= Initial Velocity of B) Substitute v^2/2g=u(v/2g-

uF=u-g(T-t) (uF= Final velocity of B at H);

uF^2=u^2-2gH

0=2v^2-2uv-2gt(u+v-gt) Solve this as v and t are known

There is a lot of correct info including the great graph. I drew the same one. I think that we should relate t2, the time t2 for projectile 2 to reach the same height that projectile 1 reaches at apogee in time t1 by t2 = k t1. Then v2 = v1/2k(1+k^2). if k=1 then v2 = v1, and as k approaches 0 v2 approaches infinity.

the only thing you need for this analysis is that t1 = v1/g and using that h1 = v1^2/2g

assume a platform in space out of the atmosphere then it would take a ball dropped of the platform t1 sec to reach v1. v1=v0+gt1 since v0 is zero t1=v1/g.

and then h1 is the same for both projectiles or dropped balls h=v2t2+1/2gt2^2

plug in t2=kt1 = kv1/g and you get the above

v2=v1/2k(1+k^2)