TARZAN

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Fahim Shahriar
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TARZAN

Unread post by Fahim Shahriar » Sun Dec 30, 2012 1:14 am

This is a question of past BdPhO.

Tarzan is hanging from a tree branch. Clayton,the gorilla hunter,takes aim at Tarzan from the same height and fires his shotgun,but Tarzan sees him and immediately lets go of the branch in order to dodge the bullet.
What happens?

(A) Tarzan is hit
(B) Tarzan dodges the bullet
(C) It depends on the speed of the bullet and distance between Tarzan and Clayton
(D) It depends on how high the branch is.
I have some confussion. Anyone explain the answer...
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Re: TARZAN

Unread post by *Mahi* » Sun Dec 30, 2012 10:20 am

(A) Tarzan is hit
Think about the bullet as a projectile, what's it's velocity along the Y-axis?
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Re: TARZAN

Unread post by SMMamun » Mon Jan 21, 2013 9:23 pm

Looking at the given answer choices, we can guess that the people who devised the question would choose (A) as its correct answer. Although (A) is better than the other three, it requires, to be correct, at least one assumption, which is not mentioned in the question. Therefore, if we critically evaluate the question—and we should critically evaluate any mathematical problem and not make any unwarranted assumption—we find that there is no correct solution to the problem. Let us elaborate:

Consider Tarzan’s case first. When he lets go off the branch, he is falling under the influence of gravity ‘in a straight-line vertically downward’ with respect to the ground. Since his initial velocity is 0, Tarzan’s flight is described by the following equation:
$h = 0.t + ½ gt^2 = 4.9t^2$ ... (1); h is the distance measured from the Tarzan’s tree branch.

So, after 1 second, Tarzan falls 4.9 meter below the tree branch. In a piece paper, try to draw the points where Tarzan would be after 2, 3, … seconds.

Now what happens to Clayton’s bullet? When the bullet leaves the shotgun, it goes in a forward motion due to the bullet’s initial forward velocity. At the same time, it simultaneously goes downward because of gravity. Both forward motion and downward motion are independent of each other and the distances that they cover in one unit of time, (say, 1 sec) are different. Thus, the bullet's path will be a continuously curved line, i.e. a parabola, not a diagonally straight line from Clayton's branch. You already know this if you have learnt the equations for forward/horizontal distance and downward/vertical distance, which are as follows:
$x = vt$ ... (2); v is the initial horizontal velocity of the bullet. v does not change over time as there is no force in the forward direction.
$y = 0.t + ½ gt^2 = 4.9t^2$ ... (3); since the bullet’s initial vertical velocity is 0, just like Tarzan.
(x, y) are measured from Clayton's branch.
In a piece of paper, try to draw the points where the bullet will be after 2, 3, … seconds. Assume a value for v, and first go along x, then along y, to plot the bullet's various positions.

Thus, after 1 sec, the bullet goes x = vt = v.1 meter forward toward Tarzan and y = 4.9 meter downward toward the ground.

So, you see that both Tarzan and the bullet are on the same horizontal level after 1 second, i.e. both are 4.9 meter down from their respective branches. Now if the tree branch is only 4.9 meter high, then the bullet will hit Tarzan only if v.1 meter ≥ the horizontal distance (let, it be d meter) between Tarzan and Clayton.
If v.1 < d, the bullet and Tarzan both fall to the ground after exactly 1 sec, but the bullet misses Tarzan.
If v.1 = d, the bullet hits Tarzan just at the ground after exactly 1 sec.
If v.1 > d, the bullet hits Tarzan above the ground and before 1 sec (because at each instant, Tarzan and the bullet are always at the same horizontal level. This is the key to understanding the question, which we would realize more below.)

Now say that the branch is much higher than 4.9 meter and the bullet, although closer to Tarzan than before, could not hit him in the 1st second. But it has still chance because their flights have not finished. Let's see what happens after 2 seconds.
Tarzan falls 19.6 meter from his branch.
The bullet goes v.2 meter forward toward Tarzan and 19.6 meter downward. Both the bullet and Tarzan are again on the same horizontal level. Now, if the branch is total 19.6 meter high, the bullet will hit Tarzan only if v.2 ≥ d, as you can guess.

At this point, the whole picture should be clear to you.
(1) at each instant, $t = ..., ¼, ½, ¾, 1, 1.5, 2, ...$ sec, whatever it might be, the bullet and Tarzan are at the same horizontal level.
(2) to be able to hit Tarzan in any of those levels, the horizontal distance covered by the bullet at that level must be equal to the fixed distance between Tarzan and Clayton. This can be ensured if v.t ≥ d.
(3) now the maximum horizontal distance that the bullet can cover in its entire flight before reaching the ground is called its range R. Thus, if R ≥ d, the bullet will catch Tarzan sooner or later. This is the implicit assumption made by the people who created the question. But R depends on both bullet speed v and branch height H, and then R needs to be compared with d. In real life, this assumption can be a good assumption given the high speed of shotgun bullets and a reasonable distance between the hunter and prey, if we forget air drag on the bullet. But why should we assume R ≥ d in this theoretical problem without further information!

Therefore, the question has no precise answer.

Note: a picture is worth a thousand words. If you draw the graphs I suggested, everything would be clear to you quickly. :)

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Re: TARZAN

Unread post by *Mahi* » Fri Jan 25, 2013 10:04 pm

SMMamun wrote:
Looking at the given answer choices, we can guess that the people who devised the question would choose (A) as its correct answer. Although (A) is better than the other three, it requires, to be correct, at least one assumption, which is not mentioned in the question. Therefore, if we critically evaluate the question—and we should critically evaluate any mathematical problem and not make any unwarranted assumption—we find that there is no correct solution to the problem. Let us elaborate:

Consider Tarzan’s case first. When he lets go off the branch, he is falling under the influence of gravity ‘in a straight-line vertically downward’ with respect to the ground. Since his initial velocity is 0, Tarzan’s flight is described by the following equation:
$h = 0.t + ½ gt^2 = 4.9t^2$ ... (1); h is the distance measured from the Tarzan’s tree branch.

So, after 1 second, Tarzan falls 4.9 meter below the tree branch. In a piece paper, try to draw the points where Tarzan would be after 2, 3, … seconds.

Now what happens to Clayton’s bullet? When the bullet leaves the shotgun, it goes in a forward motion due to the bullet’s initial forward velocity. At the same time, it simultaneously goes downward because of gravity. Both forward motion and downward motion are independent of each other and the distances that they cover in one unit of time, (say, 1 sec) are different. Thus, the bullet's path will be a continuously curved line, i.e. a parabola, not a diagonally straight line from Clayton's branch. You already know this if you have learnt the equations for forward/horizontal distance and downward/vertical distance, which are as follows:
$x = vt$ ... (2); v is the initial horizontal velocity of the bullet. v does not change over time as there is no force in the forward direction.
$y = 0.t + ½ gt^2 = 4.9t^2$ ... (3); since the bullet’s initial vertical velocity is 0, just like Tarzan.
(x, y) are measured from Clayton's branch.
In a piece of paper, try to draw the points where the bullet will be after 2, 3, … seconds. Assume a value for v, and first go along x, then along y, to plot the bullet's various positions.

Thus, after 1 sec, the bullet goes x = vt = v.1 meter forward toward Tarzan and y = 4.9 meter downward toward the ground.

So, you see that both Tarzan and the bullet are on the same horizontal level after 1 second, i.e. both are 4.9 meter down from their respective branches. Now if the tree branch is only 4.9 meter high, then the bullet will hit Tarzan only if v.1 meter ≥ the horizontal distance (let, it be d meter) between Tarzan and Clayton.
If v.1 < d, the bullet and Tarzan both fall to the ground after exactly 1 sec, but the bullet misses Tarzan.
If v.1 = d, the bullet hits Tarzan just at the ground after exactly 1 sec.
If v.1 > d, the bullet hits Tarzan above the ground and before 1 sec (because at each instant, Tarzan and the bullet are always at the same horizontal level. This is the key to understanding the question, which we would realize more below.)

Now say that the branch is much higher than 4.9 meter and the bullet, although closer to Tarzan than before, could not hit him in the 1st second. But it has still chance because their flights have not finished. Let's see what happens after 2 seconds.
Tarzan falls 19.6 meter from his branch.
The bullet goes v.2 meter forward toward Tarzan and 19.6 meter downward. Both the bullet and Tarzan are again on the same horizontal level. Now, if the branch is total 19.6 meter high, the bullet will hit Tarzan only if v.2 ≥ d, as you can guess.

At this point, the whole picture should be clear to you.
(1) at each instant, $t = ..., ¼, ½, ¾, 1, 1.5, 2, ...$ sec, whatever it might be, the bullet and Tarzan are at the same horizontal level.
(2) to be able to hit Tarzan in any of those levels, the horizontal distance covered by the bullet at that level must be equal to the fixed distance between Tarzan and Clayton. This can be ensured if v.t ≥ d.
(3) now the maximum horizontal distance that the bullet can cover in its entire flight before reaching the ground is called its range R. Thus, if R ≥ d, the bullet will catch Tarzan sooner or later. This is the implicit assumption made by the people who created the question. But R depends on both bullet speed v and branch height H, and then R needs to be compared with d. In real life, this assumption can be a good assumption given the high speed of shotgun bullets and a reasonable distance between the hunter and prey, if we forget air drag on the bullet. But why should we assume R ≥ d in this theoretical problem without further information!

Therefore, the question has no precise answer.

Note: a picture is worth a thousand words. If you draw the graphs I suggested, everything would be clear to you quickly. :)
What you said in your post, just means that option (D) is correct, if the branch is higher than needed, Tarzan is hit, and if it is not, Tarzan is saved as the bullet already bit the dust, and I think you yourself disproved the line "Therefore, the question has no precise answer.".
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Re: TARZAN

Unread post by SMMamun » Sat Jan 26, 2013 6:50 pm

@Dear Mahi:

You have either failed to read or grossly misrepresented what I actually wrote, because nowhere in my explanation did I write or mean that the branch height itself is the only parameter to determine the situation, hit vs. no-hit. I clearly wrote that the bullet will hit Tarzan only if
bullet horizontal range, R≥distance between Tarzan and Clayton, d

Then I clarified that R depends on both v and H, meaning a greater v and/or a greater H will increase the likelihood of hitting. Still I did not conclude that these are all. I finally said that R still needs to be compared with d. Thus the success of hitting ultimately depends on all three parameters: H, v, and d.
_______
It does not require much thinking to rule out the answer choices, especially (A), because (A) means that whatever the branch height, initial bullet velocity, and distance between Tarzan and Clayton may be, the bullet will always hit Tarzan. This is a ludicrous answer. It means that if Tarzan is in the Sundarbans and Clayton shoots him from Chittagong, the bullet will still hit Tarzan regardless of bullet velocity, Clayton's position from the ground, and the distance between Chittagong and the Sundarbans.

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Re: TARZAN

Unread post by sowmitra » Sun Jan 27, 2013 4:50 pm

@ SMMamun :
I think you are having trouble to see the whole picture here.

Suppose, Tarzan and the bullet are in space, and, the given scenario occurs, without any reference point in their vicinity, other than themselves. The only form of motion in them are the vertical acceleration of Tarzan and the bullet, and the horizontal motion of the bullet. Now, if you perform this thought experiment, you can see that, Tarzan and the bullet are falling at the same vertical acceleration. Since, there is no other reference point, Tarzan and the bullet would have no vertical motion relative to each other. So, the only parameter to be considered here is the horizontal velocity of the bullet. Now, as Tarzan is standing still (relative to the bullet's downward acceleration), and the bullet is coming straight at him (and he is doing nothing), Tarzan will eventually be hit.

[One major assumption of this thought experiment is that, the bullet and Tarzan will be falling forever, or at least for enough time for the bullet to reach him. So, the bullet will get ample time to reach Tarzan. If the height is finite, then, we have to calculate whether the bullet will have enough time to hit Tarzan before it falls to the ground. But, since the question is a theoretical one, the question-setters are most probably referring to the 1st condition.]

For the last part of your post......YES, the bullet WILL hit Tarzan, even if Tarzan is in the Sundarbans and Clayton shoots him from Chittagong, provided that the trees are high enough. :geek:
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Re: TARZAN

Unread post by SMMamun » Sun Jan 27, 2013 8:37 pm

@sowmitra,
Your thinking about the whole picture, and about my trouble :), is certainly wrong, because what you explained in your first major paragraph—such as only form of motion, vertical acceleration of Tarzan and bullet, horizontal motion of bullet, no relative vertical motion between them) basically provides no additional insight to our discussion but actually reiterates what I have already explained in my first comment, especially by the equation $h=0.t+½gt^2$, which, I showed, is applicable for both Tarzan and bullet. Please read my first comment to see that I have considered all these things.

You are then making a ONE MAJOR ASSUMPTION that the bullet and Tarzan will be falling forever, or at least for enough time for the bullet to reach him. But this assumption is not mentioned in the question. So, you give someone a question, but not the major assumption. What a dubious approach of questioning, just to baffle the student later! Interestingly, you are not even sure whether the question-setters were really referring to the 1st condition, because you wrote ‘most probably’. :)

When I read your last paragraph, I realize why you have failed to feel the essence of the question. You are stuck only with one parameter when you say ‘provided that the trees are high enough’ and have forgot other constraints of the problem. Let me clarify. You can imagine a sufficiently high tree, even an infinitely high tree (which is of course an unrealistic assumption) but, no problem, I can also imagine a sufficiently low bullet speed, even zero bullet speed (which is also unrealistic, but not more unrealistic than your infinitely high tree). Now make sure you can really hit Tarzan in the Sundarbans from your infinitely high tree! ;)

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Re: TARZAN

Unread post by nafistiham » Mon Jan 28, 2013 9:55 am

Well, I think more than everything of the problem has been discussed here.
But, still, I wish to add a little more.

When we say 'tarzan is hit', tarzan must be hit no matter what. Just, if i can arrange an environment that has a safe tarzan, then, we can't say that. Now, the bullet definitely has a horizontal speed, which g can never have influence on. We are trying to drop the bullet and tarzan at the same line. But, is it not dependant on the speed of the bullet, distance of calyton and tarzan ?

So, I think right answer should be
(E) (C) and (D)
:)
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