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[Dilbert]

Sorry...been out at a Lobster Festival...and I don't even like lobster. Ah, relationships...

When you and someone else you're talking with say something at the same time and they call "jinx" on you, do you automatically get to talk again since they said your name, jinx, to "jinx" you?

No, mostly because most people don't call me "Jinx" in real life. It's a nickname I earned a long time ago, but I never see the guy who dubbed me that anymore. (Thank God.) In addition to being my online nickname, it's written on my basketball as a motivational statement...other than that, it's not public knowledge.

If a tree falls in a forest, how do you know if it killed someone...

Well...I guess you'd listen for a scream. Figuring out whether the tree killed Hylas or Philnous...now THAT is a tough question.

Can you explain why some series are supposed to "converge" and some "diverge"? I was given all the rules in Calc2, but it never really made any sense to me. If you are always adding numbers to the series....all the way out to infinity.....then how can it ever converge to a single sum?

The formal definition of convergence and divergence is a pain in the ass, according to what I remember of calculus. For what it's worth, I never got a good intuition for it -- I just memorized the various tests and used them. If I'm REALLY bored tomorrow, I might open up the ol' books.

What is the best way to go about attacking physics problems?

Preferably with a heavy, sharp object.

I will add some discussion of "physics-think" in that email I owe you. It's still on my to-do list.

As for your specific problem with the two balls (oh, wait, that came out wrong): One way of attacking those problems is to think about what must be EQUAL for the two balls. They have different initial velocities, different times of flight...but they must travel the same distance (from the top to the bottom of the building), are subject to the same acceleration due to gravity, and they hit the ground at the same time. The last thing to note is that their times of flight are related -- if the first one takes t seconds to fall, the second one must take t-2 seconds to fall, since they hit at the same time.

Hopefully that gets you on the right track...

Are gay people really evil?

Nope. Some of my best friends are gay.

How many slices of pizza should I have for dinner?

Hopefully you figured this out on your own, since it's WAY past dinnertime...

How often do you shave your head?
Why are you a Giants fan?
You seem to be very confident, were you always this way? I doubt it, so, what led you to become the way you are?

Head: About once a week. I never shave it all the way down, since I think that might look a little strange on me.

Giants: I was born in Albany, NY, and spent most of my youth there. I ended up being a Giants fan, and it stuck with me ever since.

Confident: Cool, I'm fooling someone! Like a lot of people, I suspect, it's a mixed bag. There are things that I'm extremely confident about...but there are things that I'm completely insecure about. Probably the best way to gain confidence is to end up in a situation where you're pretty much totally lost...if you can find your way out, you end up feeling a lot more confident about your ability to handle ANYTHING life can throw at you.

Read my last pm?

Yup...was going to reply this weekend, time permitting. (Gotta travel for work on Sunday...YAY.)

What's the best way to, non-surgically, remove a racoon from your anus?

I only ask because this seems to be a frequent problem in the midwest.

...and they say there isn't a split between Red State/Blue State America?

Why can't I find an anti-gay gay?

I've read this question three times, and I still don't understand what you're asking...

When is enough enough?

Probably when someone starts an "official MAF backlash" thread.

Why am I so fat?

Dunno. What do you eat? How much do you exercise? How many raccoons are up your ass?

 

[Dilbert]

whatever happened to your video card

Oops, missed this one.

When I took the card out to see why the fan wasn't working, there were cracks in some of the chips -- probably from heat damage. I am now the proud owner of an X800 Pro...and I will be conducting a bake sale this weekend to pay for it. Sheesh.

 

[Dujour]

It's not my fault there're a bunch of trend following whores in here >:O

What do you want most from life?

 

[Mandark]

When I post the poll for your demodding, should it be in the Off Topic or Gaming Forum?

 

[thom]

this place by my house has all you can eat sushi...and it's actually decent.


wanna come?

 

[Wellington]

Jinx, you missed two of my questions.

 

[Loki]

Preferably with a heavy, sharp object.

I will add some discussion of "physics-think" in that email I owe you. It's still on my to-do list.

As for your specific problem with the two balls (oh, wait, that came out wrong): One way of attacking those problems is to think about what must be EQUAL for the two balls. They have different initial velocities, different times of flight...but they must travel the same distance (from the top to the bottom of the building), are subject to the same acceleration due to gravity, and they hit the ground at the same time. The last thing to note is that their times of flight are related -- if the first one takes t seconds to fall, the second one must take t-2 seconds to fall, since they hit at the same time.

Hopefully that gets you on the right track...

No sweat-- whenever you get around to it.


About the problem: like I said, I felt that (obviously) their y-displacements would have to be equivalent at some point, which would then be the minimum height of the building. This point would have to be found by using the given values of the initial velocities and the acceleration due to gravity. The thing that gets me is that not only do you do not know the height of the building (which would be their y-displacements when they were equal, considering that this is asking for the minimum possible height), but you would seemingly have to use their y-displacements to find the time of fall until they were at equal heights, or vice versa, using the time (derived somehow, likely by "equating equations" with t and t-2) to find the height at which they had equal y-displacements. In case you misread the (which I doubt-- you're probably just alluding to something I hadn't considered), the question didn't state that they hit the ground at the same point, but rather that, at some point, the ball thrown 2 seconds later with the initial velocity of 45 m/s, manages to "catch up" with the other ball dropped from rest 2 seconds earlier; if you manage to find this point in the y-coordinate plane, you'll have the minimum height of the building.


Perhaps you meant that it's more convenient to look at this as "them hitting the ground" at the same time (this is likely what you meant), and express the equivalency of their y-displacements that way? It couldn't be that once they met in midair, they'd proceed together to the ground regardless of height, because the ball thrown initially at 45 m/s would eventually overtake the other ball given enough height, correct? (via the equation v=v(i)+at ; the v(i) term for one is zero, for the other it's -45 m/s, so I'd imagine the first is going to be overtaken at some point-- just the fact that it can be thrown 2 seconds later and still "catch up" to the other ball suggests this very thing. Bah, I dunno-- I'm likely overthinking this (if there's such a thing for physics problems ). I just tried my hand at it again for a few minutes and ended up with one amazingly convoluted series of equations after another-- surely I'm doing something wrong for such a relatively simple problem, so I'll bow out.


My brain is officially turned off for the evening. Thanks for the response, and shoot me that email whenever. Any help in terms of strategies or the preferred general thought process for physics problems is welcome, not just help with this specific question.

 

[Dujour]

Is Tru Calling getting a second season?

 

[NotMSRP]

If you found a magic lamp with a genie inside, what would be your wishes in ranking order? Assume maximum of 3 wishes allowed.

 

[RobotChant]

Why can't I find an anti-gay gay?

How about I take this one.

I think your dilemma can be solved by simply talking to a Catholic priest.

 

[fart]

Is Tru Calling getting a second season?

hopefully fox will pull the plug before they're forced to air it again. FINGERS CROSSED!

 

[Phoenix]

If you found a magic lamp with a genie inside, what would be your wishes in ranking order? Assume maximum of 3 wishes allowed.

Wish 1 - infinite wishes.

Decide the rest later...

 

[Dilbert]

What do you want most from life?

As much as it's a cliché, I want to Do Great Things. The problem is that I have idea what form they will take yet. At the moment, I would hope to be remembered for my writing...but I think I'd be happy if I ended up being a good parent, when it's time for that phase of my life. Work achievements are transient at best.

When I post the poll for your demodding, should it be in the Off Topic or Gaming Forum?

Off-Topic for sure. Does anyone in the Gaming Forum even know who I am?

this place by my house has all you can eat sushi...and it's actually decent.
wanna come?

If it was anything BUT sushi, I might consider it...but I HATE sushi. Actually, I don't care for most seafood in general, but raw fish earns a special place on the "hell no" list.

Jinx, you missed two of my questions.

No I didn't. Check your PMs.

...the question didn't state that they hit the ground at the same point, but rather that, at some point, the ball thrown 2 seconds later with the initial velocity of 45 m/s, manages to "catch up" with the other ball dropped from rest 2 seconds earlier; if you manage to find this point in the y-coordinate plane, you'll have the minimum height of the building. Perhaps you meant that it's more convenient to look at this as "them hitting the ground" at the same time, and express the equivalency of their y-displacements that way?

Yup, that's exactly right. "Just catching up" means equivalency, and it relates to the minimum height of the building too. If there's no more room to fall...there's no more time to catch up.

I'd use the kinematic equation x = x0 + v0t + (1/2)at² to solve the problem. We can make the problem a little bit simpler by setting a couple of coordinate conventions in a useful way. Let's assume that the top of the building is x0 = 0, and let's also assume that going down (in the direction of gravity) is positive.

For both balls: a = 9.8 m/s², x0 = 0 m
For ball #1: v0 = 0 m/s
For ball #2: v0 = 45 m/s (positive because it's being thrown downwards)

We can write parametric equations for each ball ("x as a function of t"):

x1(t1) = 4.9(t1)²
x2(t2) = 45t + 4.9(t2)²

where t1 = the time for ball #1 and t2 = the time for ball #2. (Hard to do this without subscripts...)

We know from the problem setup that t2 = (t1 - 2), which can be rewritten as t1 = (t2 + 2) if you prefer to work with positive numbers. We also know that at some value T, x1(T) = x2(T), and the value of either function is the height of the building. So, let's rewrite the first equation to be a function of t2:

x1(t2) = 4.9(t2 + 2)²
x2(t2) = 45t + 4.9(t2)²

Substituting t2 = T (which is the catch-up time):

x1(T) = 4.9(T + 2)²
x2(T) = 45T + 4.9T²

At this point, it just becomes a math problem. As you can see, we can set the two equal to each other, and after some rearranging, you will have a function which is quadratic in T...which can always be solved using the quadratic formula.

Usually, one of the candidate roots can be rejected because it doesn't make physical sense (negative time, for example), but sometimes you actually have to accept or reject both answers. Sometimes, you even luck out and don't have to use the formula! Let's see how this one goes:

4.9(T + 2)² = 45T + 4.9T²
4.9(T² + 4T + 4) = 4.9T² + 45T
4.9² + 19.6T + 19.6 = 4.9T² + 45T

The T² terms exactly cancel!

19.6 = 25.4T
T = 1.3 sec = t2

Don't forget that we solved for t2 because of how we set up the time substitution. t1 = 3.3 seconds. Now that we have the time of flight, substitute in to either equation to get the height of the building. I'm going to use x1(t1):

x1(t1) = 4.9(t1)²
x1(3.3) = 4.9(3.3)²
x1 = 53.4 meters

Is Tru Calling getting a second season?

According to Fox's website, it premieres on November 4...which will make my friend Michelle VERY happy. She has the hots for Eliza Dushku.

If you found a magic lamp with a genie inside, what would be your wishes in ranking order? Assume maximum of 3 wishes allowed.

Assuming that my first wish for omnipotence would be granted, do I even NEED the other two?

 

[NotMSRP]

Suppose you could time travel but only once. What date would you select? As an added bonus, you could also teleport to any location on earth with the time machine. To where?

 

[Loki]

Yup, that's exactly right. "Just catching up" means equivalency, and it relates to the minimum height of the building too. If there's no more room to fall...there's no more time to catch up.

I'd use the kinematic equation x = x0 + v0t + (1/2)at² to solve the problem. We can make the problem a little bit simpler by setting a couple of coordinate conventions in a useful way. Let's assume that the top of the building is x0 = 0, and let's also assume that going down (in the direction of gravity) is positive.

For both balls: a = 9.8 m/s², x0 = 0 m
For ball #1: v0 = 0 m/s
For ball #2: v0 = 45 m/s (positive because it's being thrown downwards)

We can write parametric equations for each ball ("x as a function of t"):

x1(t1) = 4.9(t1)²
x2(t2) = 45t + 4.9(t2)²

where t1 = the time for ball #1 and t2 = the time for ball #2. (Hard to do this without subscripts...)

We know from the problem setup that t2 = (t1 - 2), which can be rewritten as t1 = (t2 + 2) if you prefer to work with positive numbers. We also know that at some value T, x1(T) = x2(T), and the value of either function is the height of the building. So, let's rewrite the first equation to be a function of t2:

x1(t2) = 4.9(t2 + 2)²
x2(t2) = 45t + 4.9(t2)²

Substituting t2 = T (which is the catch-up time):

x1(T) = 4.9(T + 2)²
x2(T) = 45T + 4.9T²

At this point, it just becomes a math problem. As you can see, we can set the two equal to each other, and after some rearranging, you will have a function which is quadratic in T...which can always be solved using the quadratic formula.

Usually, one of the candidate roots can be rejected because it doesn't make physical sense (negative time, for example), but sometimes you actually have to accept or reject both answers. Sometimes, you even luck out and don't have to use the formula! Let's see how this one goes:

4.9(T + 2)² = 45T + 4.9T²
4.9(T² + 4T + 4) = 4.9T² + 45T
4.9² + 19.6T + 19.6 = 4.9T² + 45T

The T² terms exactly cancel!

19.6 = 25.4T
T = 1.3 sec = t2

Don't forget that we solved for t2 because of how we set up the time substitution. t1 = 3.3 seconds. Now that we have the time of flight, substitute in to either equation to get the height of the building. I'm going to use x1(t1):

x1(t1) = 4.9(t1)²
x1(3.3) = 4.9(3.3)²
x1 = 53.4 meters

I know what the quadratic formula is, you rascal you. I did take math up to and including calculus, and did very well you know (though you'd never be able to tell it by my trouble with these physics problems ). Yup, as I suspected-- now that I see it done, it's a huge "DUH" moment (I've already slapped myself in the head, btw). Man, that's so simple; this goes back to what I was saying about something just not "clicking"-- it's the oddest thing. :/ I was trying to set their y-displacements equal (x in your example), but kept getting hung up because in that equation (x=1/2(v0+v)t ), there were two unknowns (time and the y-displacement) instead of just one; it ended up being much more complicated than the solution you just showed-- that was because when I tried to equate them, I kept rearranging it to solve for t (t = 2x/(v0+v) ) and then substituting v = -45 -gt for the second ball into that equation (I defined down as negative)....as you can imagine the result was a mess. That was all because I was hung up on trying to work the y-displacement into the original equations (because I figured those were the values which had to be equated), as opposed to getting t first and then using a different equation to solve for y, like you did. Gah...


I guess I'll put more time into the course (obviously), because despite my solid grade on the exam, I really don't like it when I don't feel entirely confident with the material, for any course.


Again, thank you for taking the time to lay that out like that. This doesn't absolve you of your email sending obligations, but there's no rush at all for that.


-jinx- = teh rock

And I don't know about omnipotence, but you're pretty far ahead of the pack in the "omniscience" category.

 

[Diablos]

Why do hardcore conservatives think they know everything?

 

[Comfort_Eagle]

Is love real?