Mathematics Question

Yes. Just clears the height refers to the top of the parabola.

Your point super hero throwing with great force - this view is correct and I did not think like that.

Thanks,

DHAYANANDHAN.S

If the ball is thrown at 45^o then the vertical and horizontal velocities are equal.
The system can then be solved using v=u+At, v²=u²+2AS & S=ut + 1/2At²
(EQUATIONS OF MOTION)

For the horizontal motion:
Using s=ut, (a + b) = V t therefore total flight time is t = (a+b)/V
For the Vertical motion:
Since up and down motion is symmetrical about the flagpole.
Time for the ball to reach zero vertical velocity, at max height = (a+b)/2V

Presumably you can rearrange the equations and plug in the time to get an answer.

Good luck Jeff

I don't understand why my answer is off topic.
Would whoever did it explain. Thanks Jeff

The max height of the ball doesn't really figure into this problem, even if deducible.

If you put the pole at x=0, then the ball follows a trajectory y = -Ax2 + h, i.e. an upside down parabola. The slope (derivative) is y' = -2Ax

At x = -a, y' = 1, solve for A in terms of a, A = 1/(2a)

Ball lands at x = b, y = 0.

Plug in x, y, and A into y = -Ax2 + h and solve for h

I made some plots to illustrate how I interpret the problem. The ball is thrown 'a' units from the pole (and not necessarily at ground level) at a 45-degree angle (slope = 1), passes horizontally just above the pole, and falls to the ground distance b from the pole. Given a and b, find the height, h. of the pole.

x = linear distance, y = height of ball,

equation of curve: y = h - Ax²,

slope of curve: y' = -2Ax

at x=-a, y'=1, solve A=1/(2a)

at x = b, y = 0, solve h = Ax² = Ab² = b²/2a

This doesn't fit any of your four choices, although if a=b, (the ball is thrown from ground level) option 2 would be correct.

1. (ab)/(a-b)

2. (ab)/(a+b)

3. (2ab)/(a+b)

4. (ab)/(a +2b)

So the problem isn't a math problem, but a reverse engineering problem. Figure out what problem has one of the answers above.

Please get used to the idea that the pole is not necessarily at the middle of the trajectory. (And a not necessarily equal to b.)

That said, I haven't yet ruled in option 2 as correct (versus none of the above). Still working on that part....

Actually, this is a really good problem/puzzle. Solvable with calculus, of course, but possibly without--and perhaps quite elegantly.

..."the ball just clears the pole height"...

This to me means the apex of the ball travel just clears the height of the pole, it doesn't say barely missed it...otherwise the pole height has no meaning, or definition, other than it was somewhere below the arc of the ball.....This would be like saying on my way to the store I passed a cute chick, what is the location of the girl....?.. and of the choices #2 works as a definition to boot....

You are still completely missing the point. The ball could just barely clear the pole on the way up (in which case b > a), at apex (a = b), or on the way down (a > b). Not one of your diagrams addresses this properly. This is the same point that has led the OP astray, along with a few other people.

By now at least two persons have established that only option 2 from the OP can be correct (and that in one special case it indeed is correct), but that hasn't yet proved that it is generally correct. I think so, but I haven't yet proven so, either. Thus, AFAIK to now, "none of the above" might be correct, though I doubt it.

I don't think so, a multiple choice test that has only one answer that could be correct, is going to be the answer.....could you make an argument that there could be other possibilities, yes you could...but that's not the question...

On the contrary, that is exactly the question; the OP suspects (incorrectly, I think) that the test writer got it wrong.

A ball thrown at a 45-degree angle at distance 'a' from the pole and landing distance 'b' from the pole will cross the pole at different heights, depending on the launch elevation. In other words, without specifying either the height the ball was thrown from or the slope of the trajectory above the pole, there is not a unique solution. Here are two different possible solutions:

That does not apply to the OP, which stipulates that the ball just clears the pole. Moreover, it is not given that the pole is at the center of the trajectory. Indeed, that is why there are two horizontal givens, a and b, rather than just a.

If the ball is allowed to start at an unspecified distance above the ground, then the problem is completely unsolvable, not even a formula, much less numerical answers.

If the ball is allowed to start at an unspecified distance above the ground, then the problem is completely unsolvable, not even a formula, much less numerical answers.

Actually, it is solvable if the ball is allowed to start at an unspecified distance above the ground, if the pole is at the apex of the trajectory, h = b²/2a, see #9.

A parabola is defined by three variables, so you need three independent constraints. Either the start height or the slope of the trajectory above the pole can be the third constraint to make it solvable.

The problem statement, "the ball just clears the pole height", sounds to me like that is as high as the ball goes, that the pole height is the apex. There is no mention at all of the height from which the ball is thrown, but it has to be greater than zero, unless it's a "granny shot" from the ground.

Having said all that, the fact that answer #2 works, only when a=b, leads me to believe that we are all putting more thinking into this than the problem poser intended. I'm thinking that he intended it to be a simpler problem, with the apex at the pole and the ball thrown from zero height, and not make it look too easy by revealing that a=b.

JMHO.

The real problem is translating English into Math.

I disagree vehemently, for reasons already mentioned.

I still agree with you, and there is no mention of the thrower's height, and no reason to interject details about the thrower.

a≠b remains as a most likely contribution to the range of answers. a=b is a singular result. All 45° launch angle tells us is the t=0 trajectory, as the trajectory as the ball passes over the pole can be anywhere in the subset of angles <45°, and in fact in the case of symmetry with respect to the pole, the trajectory at the pole would be 0°, the maximum height.

all we seem to know is a=h since launch angle is 45°, however, I draw attention to this fallacy of thought here: if 45° initial angle, and if maximum height is reach at distance a, then the average angle is less than 45°, in which case a>h.

I like your answer, no matter the height of the pole, the slope remains constant. Nice algebraic answer.

I don't know who gave me a 1 for off topic, but I would have felt better getting at least a 2 or 3.

This has all become a situation of one's perspective......

I agree, the pole is only in the middle of the trajectory if the ball is thrown from ground level (by a very short pitcher! ). The height from which the ball was thrown was not specified and is not necessary to solve the problem. Assuming that it is from ground level seems like a bad assumption to me.

My plot puts the pole close to the middle because in my simulation I assumed the pitcher was much shorter than the pole. And I set the origin at the pole to simplify the formula for the parabola.

However, b²/2a = ab/(a+b) --- option 2 --- only if a=b. So, maybe we're overthinking it. Maybe it's supposed to be an easier (and more unrealistic) problem than we think.

That might possibly be a point, but it is not the important point. The important point is that the pole might be far to the left or right of the ball's apex.

Indeed. There is the special case where a = h = 0 (b then depends only on the initial velocity of the ball) but all the expressions evaluate to 0, which is the correct answer.

That is NOT the problem as stated.

This diagram is the best I can do with only typography. The parabolic trajectory of the ball is represented crudely by the sloped line starting at lower left, transitioning to the horizontal line through vertex V of the parabola, and then sloping back down to the ground at a+b.

The pole, which can be anywhere along the x-axis from 0 to a+b, is shown by the vertical line from a (horizontal location) to h (vertical height).

The vertical line from (a+b)/2 to vertex V is the special case where the pole is at the center of the trajectory. Its height is (a+b)/4, but this represents only the special case. The original question is NOT ABOUT this special case; it is about the GENERAL case, where the location of the pole can vary.

Because of this, all of the drawing before this are only minimally relevant, along with most of the comments.

..................___V___ [y = (a+b)/4]
................./.......|......\
................/........|.......\
.............../.........|........\
............../h........|.........\
............./|.........|..........\
............/.|.........|...........\
.........../..|.........|............\
........../...|.........|.............\
........./....|.........|..............\
........0....a.....(a+b)/2........a+b

This seems to make an equilateral triangle

you can use Pythagoras' theorem

in this case we solve for "b"

https://en.wikipedia.org/wiki/Pythagorean_theorem

The ball could just clear the pole while on the way up, at crest, or on the way back down. Thus a and b need not be equal.

Options 1 and 4 can be rejected for asymmetry.

In the special case where the ball clears at crest, a = b and h = a/2. Thus option 3 is incorrect.

I agree with your reasoning but given that this is a maths exam question it must be soluble. Your interpretation makes the question insoluble therefore your interpretation cannot be correct. That leaves the only option as the top of the pole coinciding with the parabola maxima. A correct answer would show both the maths calculations and state the assumption made with the reason for making that assumption. If the question were in a logic examination I would not bother with the maths but just give your answer with a comment that not enough information was provided to solve the mathematics.

Wrong. I have already said that h, a, and b might not be individually soluble, but the original problem does not ask for that.

The problem may be more sophisticated than suitable for beginning math or engineering, but perfectly good as a higher math problem.

I also agree, if the ball is thrown at 45 degrees and we can ignore all drag, and other losses except gravity then the ball will travel distance "a" and if it just clears the pole it will have also traveled a height "h" equal to "a". It will also then travel an equal distance before it hits the ground again on the other side of the pole. So, "b"="a" for this case.

Therefore (2ab)/(a+b) =2a^2/(2a)=a=h

I guess this just depends on how simplified the physics of the problem becomes. If it is very simple, as in the case I suggested above where the path is essentially a triangle where the base equals 2 a and the height equals a then 3 is correct. If it is a little less simple then a parabola would be used and answer 2 is correct. Since the original question just said an angle of 45 degrees and no other information was given about the path I assumed it was to be the simplified version. Either way, it is not the end of the world.

Triangle. The pole is the base, the distance from the pole to the standing person is the side and the trajectory is the hypotenues. The angle is of no value or input

h = √a+b substitute pole height 3, distance 4, trajectory 5. = √9. On these values you will also have a right angle.

But if a=4 and h=3 then the angle would be 36.87 degrees and not 45.

1. Test designers don't always make it plain.

2. Put yourself in their place and figure out what they probably intended.

3. Apply the Holmes' Principle:

Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth

You could provide the date and address of the testing place and all of the cr4 engineers that have the time could go there and participate. When the teacher gets all of the different answers he will go crazy and retire and then your grandson will get an A for original thinking.

And what if the person throwing the ball is taller than the pole? He could be throwing it down, towards the ground on the other side of the pole.

We are, of course, ignoring wind resistance and relativistic effects for the purpose of arriving at an answer. Someone had to say it.

This is a 'maths' (mathematics) question, not a physics question. I.e., it needs a solution based on math and geometry, not on gravity and velocity.

We are looking for a solution that allows ANY POSSIBLE value for a and b.

With the information given, a few things are known:

The throw angle is 45 degrees, the pole is some distance 'a' from the thrower and the pole is some distance 'b' from where the ball hits. The problem also states that the height of the pole 'h' related to a and b by one of 4 possible formulae.

What is not known are the values of a or b, nor do we know if a and b are somehow related.

We also 'surmise' that the ball follows an arc from a to b, and since the throw angle is 45 degrees, we know that h must be less then either a or b (h<a, h<b), and that the ball 'just clears' the pole.

With this information we can 'try' different values for a and b and use these values to eliminate some of the possible answers.

So setting 'a' equal to values from 1 to 10, and likewise setting 'b' to values from 1 to 10, we can generate a range of possible solutions for h, from 'a' equal to 1/10th of 'b' to 'a' and 'b' equal or nearly equal, to 'a' being 10 times the size of 'b'.

I.e., we have the range for a in terms of b of: a = 0.1,...,0.9, 1, 1.1, ..., 10 times b.

With this we can generate tables of h for each possible solution 1 to 4.

And thus we can eliminate possible answers 1 and 3 right away.

Answer 1) is eliminated because it allows values for h that are infinitely large, thus violating h<a, h<b. It also allows nonsense negative values for h.

Answer 3) is eliminated because it allows values of h=a and h=b, thus violating h<a, h<b.

Answers 2 and 4 both allow reasonable values for h in terms of a and b. I.e., all values of h are less than a and less than b.

But in comparing answer 2 to answer 4, answer 2 provides a 'better' solution. I.e., there are solutions for 2 where h is just barely less than a or b. I.e., if a = 1 and b = 10, then h is .909, which would occur if the pole were close to point a but far from b. Under these same conditions in answer 4, h is .833, which does not satisfy (as well) the statement that the ball 'just clears' the pole.
Likewise for other values of 'a' and 'b'; in all cases, 2 provides a 'better' answer than 4.

Thus, the 'best' answer is 2.

So far, so good, but that has been established already. It has not yet been shown that option 2 is generally correct, although it works for a special case. I think it is correct, but conceivably there could be a typo in it or something that make it fail to satisfy the general case. If so, then the only correct answer would be "none of the above"; as the OP suggests.

In #39, I took your assumptions and derived option 2 for the general case...

Well done, and voted accordingly.

It is unfortunate that many irrelevancies derailed the problem for so long.

Word problems are often like that. Mathematics is precise, language is not. All we have is a statement of the problem and no chance to ask "What do you mean by this?"

Very well stated, as I seldom if ever write down anything mathematical, until I am 100% certain the problem has been fully stated, and in clear wording, and I fully understand the wording of the problem, and what the speaker intended.

That is often asking more than mathematics allows for.

Did you forget the relationship between a and b, namely that a+b meets the range equation value for displacement in the x direction?

Gee, I hope you did not expend all that effort for naught.

To brutally simplify, assume the pole tip is at the peak of the parabola. Assume also that the peak of the parabola is at elevation zero. Assume no air friction. Assume no superheroes or magic. Assume the ball has zero radius. Assume the thrower has zero height above ground. (starting to get a bit virtually magic)

The ball is thrown from point -a,-h, follows the parabola through point 0,0 and lands at point b,-h, which based on parabola geometry, b=a

dy/dx at x=-a =1

dy/dx at x=0 =0

dy/dx at x=b =-1

General formula y=cx^2

dy/dx=2*c*x at x=-a, dy/dx=1, so c=-1/(2*a)

y=(-1/(2*a))*x^2

At x=-a, y=-a/2 So, h=a/2

Any formula where a=b and reduces to h=a/2 is correct.

Since when is a rigorous mathematical analysis off topic?

Assumption of symmetric trajectory over the pole. My bad, so I removed the OT.

That was his bad, not yours. You have admitted to, and then incorrectly "retracted", an action that wasn't yours in the first place.

What you gave was not a rigorous analysis. It addressed only a special case and ignored the general case.

Your concluding statement is completely ridiculous. All sorts of formulas might be given that satisfy the special case but fail the general case.

The test question IS a special case and the assumptions are necessary to arrive at a single answer. Perhaps the test question would have been better if it stated that the flagpole was in Phoenix, the thrower was an average little league pitcher, that the ball just clears the tip of the flagpole at it's highest point of arc and to neglect air friction. It is clear from the question that the tester is fishing for a parabola with defined starting and midpoint boundary conditions.

In any case I suppose a meaningful answer with defined assumptions is better than, "I guess it depends..." which from an engineering standpoint is just slightly short of useless unless you want a range of height from zero to infinity minus 1.

I haven't seen a treatment yet including atmospheric pressure lapse rate, front surface ablation pressure decelleration or decrease in gravity constant with radius, although someone did mention escape velocity. I suppose at some point reality has to intrude, unless you want to get into whether all werewolves can swim.

Wrong! The test question is most emphatically NOT just the special case. Moreover, it is perfectly well formulated.

Just a subtle point.

The question has multiple choice answers: If the correct answer is for all a and b then it must be correct for a=b.

2.) is the only answer which is correct for a=b

So it must be the correct answer. That would be the easiest way to solve it if you were in the exam. Especially if you were studying for that sort of question you would probably be well aware of the special case: maximum range (on flat ground) for a given launch velocity is achieved by launching at 45°, and, the height reached in that case is ¼ the range.

That does not follow until proven. Conceivably (2) could have been incorrect when a ≠ b. That was the whole point of the OP's question--that (2) might also be incorrect. (I.e., a none-of-the-above situation.)

By now, of course, Rixter has proven for the general case that (2) is correct.

But the OP's reasoning was based on the bizarre "belief" that the ball flew in a straight line from the launch point to the max point. Surely we discounted that idea and went back to the idea that the examiners knew what they were doing.

I don't agree that that characterizes the matter correctly. In particular, the OP almost immediately calls the ball's path a parabola, not a straight line.

"Since the angle of throw is 45 Deg., the height of the pole becomes opposite side of the triangle and the distance between the person and pole becomes adjacent side of the triangle.

In the Trigonometry opposite side/adjacent side = tan 45 = 1.

Hence the height of the pole is equal to the distance between the pole and the person."

You're right--after saying parabola, the OP goes on to contradict himself.

a=b is the special case. a=h occurs how (which is the key aspect of the original statement). Last time I checked, a launch at 45 degrees (or any other angle) only travels in a straight line in space removed from the acceleration of gravity.

Not sure what you mean by a=h. For the general parabola h=a/2.

BTW Pete and Tornado walked across a bridge. Pete fell off......

No. The problem does not state the energy, nor does it state the ball never travels higher than the pole (but you assumed this in your considerations).

Indeed, if the ball is thrown at distance a, at 45° angle, and just barely misses the top of the pole (or infinitesimally grazes the pole), this ball must still follow the laws of basic classical physics. All we know is that point a is directly related to h, but that b can be quite farther out. We know the sum of a+b is the range.

Look at this again with your range equation, and come up with something to substitute in to arrive at a, then you will have h, and also b.

2 is the answer, as usual.

The path is a parabola, as others have said, but I don't know whether you're allowed to assume that. Strictly speaking you would have to show it.

But assuming it is a parabola, and that one of the choices is correct, the answer must be symmetrical in a and b (one for the ball rising, the other falling). That rules out 1. and 4.

Putting a = b, 3. gives height = a. For 45 deg launch angle, by simple geometry the ball would rise to height a only if there were no gravity. So in practice the height is < a and the answer must be 2.

There's a bit of hindsight in that. I got in a tangle with the maths but Rixter's #39 covers it nicely (still assuming a parabola). Another GA from me!