Projectiles: Newsletter Challenge (04/01/08)

First, we assume, as it does not state otherwise, that the rockets are launched veritcally. There is no cause to believe that there is a downrange component in the velocity.

Next, we need to find the time it takes rocket 1 to get to its maximum altitude. This will help us to know how long rocket 2 took to get there. We will call its maximum altitude (or height) h. We use the equation v = v_o - gt where v_o is its initial velocity. Since velocity at max is zero, this becomes 0 = v_o - gt. Solving for t we get the time it takes to get to maximum altitude and therefore have zero velocity:

t = v_o/g but the problem gives its initial velocity as v, so we will substitute that in.

t = v/g

This is how long it takes rocket 1 to get to height h. Rocket 2 arrives at height h at the same time, so rocket 2 takes time = v/g -t since it was launched at time t after rocket 1.

Now let's look at rocket 2s ascent. We will use the equation of the form

Δy = h = v_ot -½gt²

This time we solve for v_o or rocket 2's initial velocity.

v_o = (h + ½gt²)/t

In this case, t is really v/g - t, so let's make that substitution.

v_o = (h + ½g(v/g - t)²)/(v/g - t)

This is a little messy, so we do some algebra which results in the simpler form:

v_o = (gh + ½(v-gt)²)/(v-gt)

There, now everyone can get some rest.

C'mon Prof - not only does the challenge "not state otherwise" but the first sentence specifically states that the projectiles are launched vertically.
Then, we don't need the parameter 'h' in the answer - so finish the job.

I hope your instructions to students include at least the following:
1) RTFQ
2) So far as is possible, present the answer in terms of the variables given in the question.

Picky picky. Prejudiced against h are we? OK. If you insist, the equivalent equation is:

v_o = (v² + (v-gt)²)/(2(v-gt))

So go home and get some sleep. Tomorrow will be a better day.

PS. Some of the discussions were sounding rather sideways.

No prejudice against Planck's constant - it just wasn't relevant to the question

If you are going to spend any time in CR4, you will just have to get used to the ignorant dominating the space for long after a proper answer is presented, or you will go mad. As a corollary, there is no point expanding on existing answers unless you are prepared to be (even?) more meticulous than they were.

"Tomorrow will be a better day" Dubious

Lovely elegant question.

Should have specified identical

projectiles re mass and air resistance.

Two projectiles are launched vertically. Projectile A is launched with an initial velocity v. Projectile B is launched t seconds after the first. Projectile B passed Projectile A as the latter reached its highest point. Determine the initial velocity of projectile B.

Solution neglecting wind resistance:

initial velocity of projectile B = Vbo

ta = time in seconds of A projectile to reach apex.

t = time in seconds B projectile launches after A launches.

Vbo = 16[ 2(ta)^2 - 2(ta)t + t^2] / ((ta)-t)

The above is the correct answer for this problem. Note t<ta.

Are you one of PhysicsProf's students? He didn't answer the question first time around either.

No sir, I'm just an instrument tech.

That was just meant to say that, accurate or not, the data used in your solution were not those in the challenge - so you haven't (yet) answered it. The prof did the same first time around - I'm not sure whether you would regard yourself as being in good company...

Solution to Projectile Problem: J

Earth's gravitational acceleration on surface of the planet at sea level = 32ft/s².

Direction is considered downward, a negative value if positive values are upward. Therefore, acceleration of the g-field = -32ft/s² = a = acceleration.

Projectile motion equations for single dimension shot in the vertical direction are as follows: at + V₀ = V and 1/2at² + V₀ t + S₀ = S. V₀ = initial velocity. S₀ = Distance reference at Starting Point. S = final distance. Distance spanned = (S - S₀)

Set up equations of projectiles spanning the same distance meeting at the apex of projectile A: (½)a(t_A) ² + V_A0 t_A = S- S₀ = (½)a(t_B) ² + V_B0 t_B .

Therefore, (½)a(t_A) ² + V_A0 t_A = (½)a(t_B) ² + V_B0 t_B

Note that the condition of the problem states that (S- S₀) is equivalent to both projectile paths only when projectile A has reached its APEX. Another bit of valuable information.

When the APEX is reached the V=0 for the following equation: at + V₀ = V = 0. Since a = -32ft/s² , we substitute as follows: -32t_A+ V_A0 = V = 0. -32t_A+ V_A0 = 0 _.

32t_A = V_A0 . Now that we have confirmed the initial velocity of projectile A, we substitute V_A0 = 32t_A into the top bold faced equation:

(½)a(t_A) ² + V_A0 t_A = (½)a(t_B) ² + V_B0 t_B becomes

(½)a(t_A) ² + (32t_A) t_A = (½)a(t_B) ² + V_B0 t_B

(½)a(t_A) ² + 32 t_A² = (½)a(t_B) ² + V_B0 t_B note a = -32ft/s².

(½)(-32)(t_A) ² + 32 t_A² = (½)(-32)(t_B) ² + V_B0 t_B

(-16)(t_A) ² + 32 t_A² = (-16)(t_B) ² + V_B0 t_B Noting that t_{B =} t_A-t

(-16)(t_A) ² + 32 t_A² = (-16)( t_A-t) ² + V_B0 (t_A-t)

16 t_A² = (-16)( t_A-t) ² + V_B0 (t_A-t)

16 t_A² = (-16)( t_A²-2t_At + t²) + V_B0 (t_A-t)

16 t_A² = (-16 t_A² + 32 t_At -16 t²) + V_B0 (t_A-t)

32 t_A² - 32 t_At + 16 t² = V_B0 (t_A-t)

V_B0 = (32 t_A² - 32 t_At + 16 t²) / (t_A-t)

V_B0 = 16 (2 t_A² - 2 t_At + t²) / (t_A-t) This is the final answer such that {t < t_A }.

You may choose values for t_A and t at your discretion such that {t < t_A }. Plug the values that you wish to solve for the initial velocity of projectile B. This is the complete and accurate solution to the problem.

apart from 32ft/sec² not necessarily being the appropriate units, and certainly not exact, the problem gives v (I would prefer v_A) as the input data. I know substitution is 'trivial', but if you want to lay claim to providing the complete ... solution to the challenge, the answer really needs to be in a compatible format.

Strangely, I get the initial velocity of B (when T = 0) as zero.

Too much lateral thinking?

Changing the question while thinking, perhaps?
The projectiles must pass when A is at its top point. Zero velocity would be a solution for them passing when projectile B is at its top point.
BTW, that could be regarded as a more 'interesting' case, as in general there are either two answers or none.

Projectile A will impact ground zero before B will impact the zero level. Therefore, only one answer exists for this problem. Note, both projectiles are subject to the same dominant gravitational force established by the GEOID. This problem assumes that neither projectile attains the escape velocity of the planet. I assumed planet earth's gravitational field and used a constant to serve as the local acceleration over a minimal trajectory. The published solution should verify this reasoning.

Perhaps I was not sufficiently clear - I was writing about the modified problem in which projectile B is launched so that the projectiles pass when projectile B is at the top of its trajectory. As shown by BobD and followers, the original question has a unique solution.

If A projectile passes B projectile, re-label "A" to "B" and "B" to "A" and the equation holds true, again.

Yes, the equations hold true (so long as we remember also to change the sign of TB).

But the equation is no longer explicit. We need to solve the quadratic to find VB. The result is that not all combinations of TB and VA give solutions. I think you will find that if you launch B before A there will either be two solutions (both with upwards velocity) or none.

"...there will either be two solutions..."

OK, maybe it's so simple I just can't see it, so lend me some glasses before I make a spectacle of myself, please. But wouldn't one of those two solutions have to follow a negative timeline? <-blind...

This is for launching B before A, knowing when and how fast A will be launched - and requiring that A passes B when B is at its highest point (heavens knows why you would wish to do this, but it's taken from the implication of MadMike's comment).
The obvious case is when they are launched at the same time - B can be stationary or at the same velocity as A. The easiest way to see this might be to consider when B would need to be launched for A to pass it at a given height. One way to show the effect would be to draw a pair of identical parabolas (like MPM's), one on a sheet of paper (for projectile A) with a base-line representing the launch height, and one on a transparency (for projectile B). If you place the transparency with its parabola's highest point coincident with the launch of projectile A, and then slide the transparency up the parabola, you will see the launch point for projectile B starts coincident with projectile A, gets earlier for a while and then gets later. [There's only one valid solution if you launch B after A, unfortunately]

Alternatively, you could solve BobD's equation with the velocity variables swapped...

See message #120

Two days after the Official Answer was supposed to be posted, and it hasn't shown up yet. No one else has even commented on the delay. Has anyone noticed? Or has everyone stopped caring?

Inquiring minds want to know! (At least one does, anyway.)

Personally, I don't care, as I can't see that it will add anything: BobD got his sums right, and PhysicsProf among others (eventually) provided a worked version.

So you're saying "stick a fork in it"?

Where is the answer?

Where is the official answer?

The answer is posted.