# Newton’s Puzzle

#### chisigma

##### Well-known member
Dear friends of MHB

Yesterday my attention has been attracted by the following article…

http://www.huffingtonpost.com/2012/05/27/shourryya-ray-16-year-old_n_1549172.html

It seems that a 16 old student named Shouryya Ray and ‘dubbed Genius’ has ‘cracked’ a puzzle that Isaac Newton proposed 350 years ago and that nobody solved till now. Ray says that the use of the computer had an essential role in his ‘memorable enterprise'. The ‘Newton’s Puzzle’ seems to be to find a procedure to arrive at exact calculations of a trajectory under gravity and subject to air resistance

Very well !... it is well known the fact that journalist often combine ‘exaltation of myths’ and poor knowledge of specific arguments, so that probably the article has to be read carefully and many statements put in right prospective. For example it seems that such type of problems has never been seriously studied and that is of course is a nonsense. The air resistance plays an essential role in the motion of a projectile of artillery and artillery did exist also at the times of Romans … and even before!...

Anyway such type of problem is interesting and actual also in the 21-th century, so that some year ago I found practically the procedure to arrive to the solution of this problem and the computer didn’t play any special role in that. About the ‘motivation’ that pushed me to ‘attack’ this problem we will discuss later . Before showing You the procedure I used, it would be interesting some comments from You…

Kind regards

$\chi$ $\sigma$

#### Ackbach

##### Indicium Physicus
Staff member
If you search for the boy's name on Google, the only thing that comes up is a repetition of the story to which you linked. In particular, a search on Google Scholar reveals no paper that he has written. Therefore, there does not seem to be a way actually to examine his solution. Until that happens, I can't say one way or the other.

#### CaptainBlack

##### Well-known member
Covered in Aperiodical >>here<< still not much definite information about what has been done.

CB

#### chisigma

##### Well-known member
Very well!... waiting for news, let me show how I organized at the time the solution of the problem. The main hypothesis is that a meterial point of mass m having speed $\overrightarrow {v}= v_{x}\ \overrightarrow {i} + v_{y}\ \overrightarrow {j}$ is subject to an air resistance $\overrightarrow{a}= a_{x}\ \overrightarrow{i} + a_{y}\ \overrightarrow {j}$. Now we suppose to be in a two dimension cartesian system where th force of gravity is directed to the bottom. so that the components $v_{x}(t)$ and $v_{y}(t)$ of the velocity of the material point must obey to the differential equations...

$\displaystyle v_{x}^{\ '} = -k\ v_{x}^{2}\ \text{sgn} (v_{x})$ (1)

$\displaystyle v_{y}^{\ '}= -g -k\ v_{y}^{2}\ \text{sgn} (v_{y})$ (2)

... where $\displaystyle g= 9.81 \frac{m}{s^{2}}$ is the gravity acceleration and k is the so called 'air resistance coefficient' that dimensionally is measured in $m^{-1}$. The solution of (1) that I propose holds for $x\ge 0$ and $x^{\ '}\ge 0$ , but that is not a limitation...

$\displaystyle v^{\ '}_{x} = -k\ v^{2}_{x}$ (3}

The (3) is 'with separable variables' and its solution is 'easy' ...

$\displaystyle \frac{dv_{x}}{v_{x}^{2}}= -k\ dt \implies \frac{1}{v_{x}}= k\ t + \frac{1}{v_{x}(0)} \implies v_{x}= \frac{v_{x}(0)}{1+v_{x}(0)\ k\ t}$ (4)

Integrating (4) we can find x(t) as follows...

$\displaystyle x= x_{0}+ \frac{1}{k}\ \ln \{1+ v_{x}(0)\ k\ t \}$ (5)

The first part is completed... but the next part is quite more difficult and requires a dedicated post...

Kind regards

$\chi$ $\sigma$

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#### Ackbach

##### Indicium Physicus
Staff member
Isn't $|v|^{2}=\dot{x}^{2}+\dot{y}^{2}$?

#### chisigma

##### Well-known member
Isn't $|v|^{2}=\dot{x}^{2}+\dot{y}^{2}$?
Yes, of course!... the 'core' of the solution is that in the x and y directions acts an impeding force proportional to the square of the velocity component in x and y directions...

Kind regards

$\chi$ $\sigma$

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#### Ackbach

##### Indicium Physicus
Staff member
So, let me start from first principles. We describe the projectile motion in two dimensions, $\hat{i}$ and $\hat{j}$. Its position we write as ${\bf r}=\langle x,y\rangle$. We'll let $\hat{j}$ point positive upwards. The only forces on the projectile are gravity and the quadratic air resistance. The velocity is ${\bf v}=\dot{{\bf r}}=\langle \dot{x},\dot{y}\rangle$, and the acceleration is ${\bf a}=\dot{{\bf v}}=\ddot{{\bf r}}=\langle \ddot{x},\ddot{y}\rangle$. The speed is the magnitude of velocity $v=|{\bf v}|=\sqrt{\dot{x}^{2}+\dot{y}^{2}}$, as mentioned before. The projectile's mass is $m$, and we assume the air resistance force is quadratic in the speed, and in the direction opposite that of the velocity. Hence, we can write the force ${\bf F}$ as
$${\bf F}=-mg\hat{j}-kv^{2}\,\frac{{\bf v}}{v}=-mg\hat{j}-kv{\bf v}.$$
Notice that I used ${\bf v}/v$ as a unit vector in the direction of the velocity.

The momentum we write as ${\bf p}$, and for a constant mass, ${\bf p}=m{\bf v}$. Hence, Newton's Second Law dictates that
$$\dot{{\bf p}}=m\dot{{\bf v}}=m\ddot{{\bf r}}={\bf F}=-mg\hat{j}-kv{\bf v}=-mg\hat{j}-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{{\bf r}}.$$
We take this form:
$$m\ddot{{\bf r}}=-mg\hat{j}-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{{\bf r}},$$
and reduce to components thus:
\begin{align}
m\ddot{x}&=-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{x}\\
m\ddot{y}&=-mg-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{y}.
\end{align}

These equations do not decouple the way you've outlined, unless I'm missing something very basic. They do admit of one integration immediately in the usual way.

#### chisigma

##### Well-known member
So, let me start from first principles. We describe the projectile motion in two dimensions, $\hat{i}$ and $\hat{j}$. Its position we write as ${\bf r}=\langle x,y\rangle$. We'll let $\hat{j}$ point positive upwards. The only forces on the projectile are gravity and the quadratic air resistance. The velocity is ${\bf v}=\dot{{\bf r}}=\langle \dot{x},\dot{y}\rangle$, and the acceleration is ${\bf a}=\dot{{\bf v}}=\ddot{{\bf r}}=\langle \ddot{x},\ddot{y}\rangle$. The speed is the magnitude of velocity $v=|{\bf v}|=\sqrt{\dot{x}^{2}+\dot{y}^{2}}$, as mentioned before. The projectile's mass is $m$, and we assume the air resistance force is quadratic in the speed, and in the direction opposite that of the velocity. Hence, we can write the force ${\bf F}$ as
$${\bf F}=-mg\hat{j}-kv^{2}\,\frac{{\bf v}}{v}=-mg\hat{j}-kv{\bf v}.$$
Notice that I used ${\bf v}/v$ as a unit vector in the direction of the velocity.

The momentum we write as ${\bf p}$, and for a constant mass, ${\bf p}=m{\bf v}$. Hence, Newton's Second Law dictates that
$$\dot{{\bf p}}=m\dot{{\bf v}}=m\ddot{{\bf r}}={\bf F}=-mg\hat{j}-kv{\bf v}=-mg\hat{j}-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{{\bf r}}.$$
We take this form:
$$m\ddot{{\bf r}}=-mg\hat{j}-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{{\bf r}},$$
and reduce to components thus:
\begin{align}
m\ddot{x}&=-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{x}\\
m\ddot{y}&=-mg-k\sqrt{\dot{x}^{2}+\dot{y}^{2}}\,\dot{y}.
\end{align}

These equations do not decouple the way you've outlined, unless I'm missing something very basic. They do admit of one integration immediately in the usual way.
For simplicity sake lets eliminate the gravity and consider a material point having velocity $\overrightarrow{v}= v_{x}\ \overrightarrow {i} + v_{y}\ \overrightarrow {j}$ and subject only to air resistance $\overrightarrow {a}= a_{x}\ \overrightarrow {i} + a_{y}\ \overrightarrow {j}$. Now is very intuitive that if $\overrightarrow {v}$ has only the component along the x axis, then the same will be also for the $\overrightarrow {a}$ and vice versa. But that means that each component of $\overrightarrow {a}$ depends only from the corresponding component of $\overrightarrow {v}$. On the basis of this consideration the equations describing a meterial point subject only to air resistance are...

$\displaystyle v_{x}^{\ '} = -k\ v_{x}^{2}\ \text{sgn} (v_{x})$

$\displaystyle v_{y}^{\ '} = -k\ v_{y}^{2}\ \text{sgn} (v_{y})$

I think that [probably...] not all You will agree with me and some more discussion will be unavoidable, so that I suggest to separate the explanation on my solution of some years ago from the discussion about the validity of the hypotheses on which my solution is based. Is it acceptable?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
For simplicity sake lets eliminate the gravity and consider a material point having velocity $\overrightarrow{v}= v_{x}\ \overrightarrow {i} + v_{y}\ \overrightarrow {j}$ and subject only to air resistance $\overrightarrow {a}= a_{x}\ \overrightarrow {i} + a_{y}\ \overrightarrow {j}$. Now is very intuitive that if $\overrightarrow {v}$ has only the component along the x axis, then the same will be also for the $\overrightarrow {a}$ and vice versa. But that means that each component of $\overrightarrow {a}$ depends only from the corresponding component of $\overrightarrow {v}$. On the basis of this consideration the equations describing a meterial point subject only to air resistance are...

$\displaystyle v_{x}^{\ '} = -k\ v_{x}^{2}\ \text{sgn} (v_{x})$

$\displaystyle v_{y}^{\ '} = -k\ v_{y}^{2}\ \text{sgn} (v_{y})$

I think that [probably...] not all You will agree with me and some more discussion will be unavoidable, so that I suggest to separate the explanation on my solution of some years ago from the discussion about the validity of the hypotheses on which my solution is based. Is it acceptable?...

Kind regards

$\chi$ $\sigma$
We have vectors here, the air resistance is proportional to the square of the speed and directed in the opposite direction to the velovity :

$${\bf{v}}'= - k\;|{\bf{v}}|^2 \; \widehat{{\bf{v}}} = - k\; |{\bf{v}}| \; {\bf{v}}$$

so in component form:

$$v_x' = - k\; \sqrt{v_x^2+v_y^2}\; v_x$$

$$v_y' = - k\; \sqrt{v_x^2+v_y^2}\; v_y$$

You need this form when you include gravity as gravity sets a preferred direction.

CB

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#### Ackbach

##### Indicium Physicus
Staff member
Now is very intuitive that if $\overrightarrow {v}$ has only the component along the x axis, then the same will be also for the $\overrightarrow {a}$ and vice versa. But that means that each component of $\overrightarrow {a}$ depends only from the corresponding component of $\overrightarrow {v}$.
But you're over-generalizing. You can't argue from two special cases to all cases. I would agree that if the velocity happens to be in any of the $-\hat{i}, \hat{i}, -\hat{j}, \hat{j}$ directions, then the air resistance will be, respectively, in the $\hat{i}, -\hat{i}, \hat{j}, -\hat{j}$ directions. I would agree that the direction of the air resistance in one component is dependent only on the one component. But the magnitude of the air resistance depends on both components. You simply cannot ignore one component in computing the magnitude of the air resistance.

#### chisigma

##### Well-known member
Very well dear friends!... I'm an 'old wolf' and it was not difficult for me to foresee Your 'critics'... but, because, 'wolf or not wolf', I'm old I strongly beg You to leave me the chance to face a challenge one by one, that means first to complete my post #4 and after to discuss the general problem of a moving material point subject to air resistance... in any case what I will write in my next post is valid for a material point constrained on the y axes and subject to gravity and air resistance...

I'm sure that all You will give me the chance...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
... the main hypothesis is that a meterial point of mass m having speed $\overrightarrow {v}= v_{x}\ \overrightarrow {i} + v_{y}\ \overrightarrow {j}$ is subject to an air resistance $\overrightarrow{a}= a_{x}\ \overrightarrow{i} + a_{y}\ \overrightarrow {j}$. Now we suppose to be in a two dimension cartesian system where th force of gravity is directed to the bottom. so that the components $v_{x}(t)$ and $v_{y}(t)$ of the velocity of the material point must obey to the differential equations...

$\displaystyle v_{x}^{\ '} = -k\ v_{x}^{2}\ \text{sgn} (v_{x})$ (1)

$\displaystyle v_{y}^{\ '}= -g -k\ v_{y}^{2}\ \text{sgn} (v_{y})$ (2)

... where $\displaystyle g= 9.81 \frac{m}{s^{2}}$ is the gravity acceleration and k is the so called 'air resistance coefficient' that dimensionally is measured in $m^{-1}$. The solution of (1) that I propose holds for $x\ge 0$ and $x^{\ '}\ge 0$ , but that is not a limitation...
The solution of the second DE, that we now write again…

$\displaystyle v_{y}^{\ '}= -g -k\ v_{y}^{2}\ \text{sgn} (v_{y})$ (1)

… is quite more complex because the presence of the sign function cannot be ignored and that forces to consider separately the two cases $v_{y}>0$ and $v_{y} \le 0$. In the case $v_{y}>0$ the (1) becomes...

$\displaystyle v_{y}^{\ '}= -g -k\ v_{y}^{2}$ (2)

Its solution in relatively 'easy' because the variables can be separated...

$\displaystyle \frac{d v_{y}}{g+k\ v_{y}^{2}}= - dt \implies \tan^{-1} (\sqrt{\frac{k}{g}}\ v_{y} )= - \sqrt{g\ k}\ t + c \implies \tan^{-1} \{ \sqrt{\frac{k}{g}}\ (v_{y}-v_{y}(0))\}= - \sqrt{g\ k}\ t \implies v_{y}= v_{y}(0) - \sqrt{\frac{g}{k}}\ \tan (\sqrt{g\ k}\ t)$ (3)

The ordinate y of the material point is obtained integrating (3)...

$\displaystyle y=y_{0} + v_{y} (0)\ t + \frac{1}{k}\ \ln \cos (\sqrt{g\ k}\ t)$ (4)

In the case $v_{y} \le 0$ the (1) becomes...

$\displaystyle v_{y}^{\ '}= -g +k\ v_{y}^{2}$ (5)

... and the solution is similar to the previous case...

$\displaystyle \frac{d v_{y}}{-g+k\ v_{y}^{2}}= dt \implies \coth^{-1} (\sqrt{\frac{k}{g}}\ v_{y} )= - \sqrt{\frac{g}{k}}\ t + c \implies \coth^{-1} \{ \sqrt{\frac{k}{g}}\ (v_{y}-v_{y}(0))\}= - \sqrt{g\ k}\ t \implies v_{y}= v_{y}(0) - \sqrt{\frac{g}{k}}\ \coth (\sqrt{g\ k}\ t)$ (6)

... and again the ordinate y is obtained integrating (6)...

$\displaystyle y=y_{0} + v_{y} (0)\ t - \frac{1}{k}\ \ln \sinh (\sqrt{g\ k}\ t)$ (7)

Pratically the traiectory of the material point is found in two successive phases, the first with $v_{y}>0$ and the second with $v_{y} \le 0$...

Kind regards

$\chi$ $\sigma$

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#### chisigma

##### Well-known member
Let's now come back to the problem of the motion of a material point subject only to the air resistance [for example an air balloon on which the aerostatic push neutralizes the gravity...], supposing to be in two dimensions. If x(t) and y(t) are the coordinates of the material point, its polar coordinates are by definition...

$\displaystyle x= \rho\ \cos \theta$

$\displaystyle y=\rho\ \sin \theta$ (1)

First and essential hypothesis to be done is that the nature of the problem is independent from the origin of the coordinates system and from rotation, i.e. at the time t=0 we are free to set x(0)=y(0)=0 and define the vector $\overrightarrow {v}(0)= v_{x}(0)\ \overrightarrow {i} + v_{y}(0)\ \overrightarrow {j}$. As we said before the problem is independent from the rotation and that means that $\theta$ is constant in the time, so that from (1) we derive...

$\displaystyle x^{\ '}= v_{x} = \rho^{\ '}\ \cos \theta$

$\displaystyle y^{\ '}= v_{y} = \rho^{\ '}\ \sin \theta$ (2)

... and ...

$\displaystyle x^{\ ''}= v^{\ '}_{x} = \rho^{\ ''}\ \cos \theta$

$\displaystyle y^{\ ''}= v^{\ '}_{y} = \rho^{\ ''}\ \sin \theta$ (3)

At this point it is quite obvious that the motion equation involves only $\rho$ and, setting $w = \rho^{\ '}$ is...

$\displaystyle w^{\ '}= -k\ w^{2}\ \text{sgn} (w)$ (4)

Now from (2) we derive...

$\displaystyle \rho^{\ '}=w= \frac{v_{x}}{\cos \theta}=\frac{v_{y}}{\sin \theta}$ (5)

... so that (4) is equivalent to the pair of equations...

$\displaystyle v^{\ '}_{x}= -k_{x}\ v^{2}_{x}\ \text{sgn} (v_{x})\ ,\ k_{x}= \frac{k}{\cos \theta}$

$\displaystyle v^{\ '}_{y}= -k_{y}\ v^{2}_{y}\ \text{sgn} (v_{y})\ ,\ k_{y}= \frac{k}{\sin \theta}$ (6)

Honestly the result expressed by the (6) is, at least for me, quite surprising. My hypothesis that the velocities along the x axis and along the y axis can be treated separately is correct... but, and that is the 'surprise', the air resistance coefficients along the x and y axes are different ... and that changes strongly the terms of the problem...

At this point some comment from You will be very appreciated...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Let's now come back to the problem of the motion of a material point subject only to the air resistance [for example an air balloon on which the aerostatic push neutralizes the gravity...],
Air drag proportional to the square of speed only applies at high Reynolds numbers, a balloon will usually be operating at low Reynolds numbers where drag is more nearly proportional to the speed.

CB

#### CaptainBlack

##### Well-known member
Let's now come back to the problem of the motion of a material point subject only to the air resistance [for example an air balloon on which the aerostatic push neutralizes the gravity...], supposing to be in two dimensions. If x(t) and y(t) are the coordinates of the material point, its polar coordinates are by definition...

$\displaystyle x= \rho\ \cos \theta$

$\displaystyle y=\rho\ \sin \theta$ (1)

First and essential hypothesis to be done is that the nature of the problem is independent from the origin of the coordinates system and from rotation, i.e. at the time t=0 we are free to set x(0)=y(0)=0 and define the vector $\overrightarrow {v}(0)= v_{x}(0)\ \overrightarrow {i} + v_{y}(0)\ \overrightarrow {j}$. As we said before the problem is independent from the rotation and that means that $\theta$ is constant in the time, so that from (1) we derive...

$\displaystyle x^{\ '}= v_{x} = \rho^{\ '}\ \cos \theta$

$\displaystyle y^{\ '}= v_{y} = \rho^{\ '}\ \sin \theta$ (2)

... and ...

$\displaystyle x^{\ ''}= v^{\ '}_{x} = \rho^{\ ''}\ \cos \theta$

$\displaystyle y^{\ ''}= v^{\ '}_{y} = \rho^{\ ''}\ \sin \theta$ (3)

At this point it is quite obvious that the motion equation involves only $\rho$ and, setting $w = \rho^{\ '}$ is...

$\displaystyle w^{\ '}= -k\ w^{2}\ \text{sgn} (w)$ (4)

Now from (2) we derive...

$\displaystyle \rho^{\ '}=w= \frac{v_{x}}{\cos \theta}=\frac{v_{y}}{\sin \theta}$ (5)

... so that (4) is equivalent to the pair of equations...

$\displaystyle v^{\ '}_{x}= -k_{x}\ v^{2}_{x}\ \text{sgn} (v_{x})\ ,\ k_{x}= \frac{k}{\cos \theta}$

$\displaystyle v^{\ '}_{y}= -k_{y}\ v^{2}_{y}\ \text{sgn} (v_{y})\ ,\ k_{y}= \frac{k}{\sin \theta}$ (6)

Honestly the result expressed by the (6) is, at least for me, quite surprising. My hypothesis that the velocities along the x axis and along the y axis can be treated separately is correct... but, and that is the 'surprise', the air resistance coefficients along the x and y axes are different ... and that changes strongly the terms of the problem...

At this point some comment from You will be very appreciated...

Kind regards

$\chi$ $\sigma$
This is trivial, in the absence of gravity the motion is confined to 1-dimension, it is along the line of the initial velocity. Having already solved the problem with initial velocity along a co-ordinate axis this follows by a simple rotation of the space.

CB

#### CaptainBlack

##### Well-known member
But you're over-generalizing. You can't argue from two special cases to all cases. I would agree that if the velocity happens to be in any of the $-\hat{i}, \hat{i}, -\hat{j}, \hat{j}$ directions, then the air resistance will be, respectively, in the $\hat{i}, -\hat{i}, \hat{j}, -\hat{j}$ directions. I would agree that the direction of the air resistance in one component is dependent only on the one component. But the magnitude of the air resistance depends on both components. You simply cannot ignore one component in computing the magnitude of the air resistance.
He probably can in the case where he ignores gravity, since you can just solve the problem in the frame with the x-axis along the direction of the initial velocity, and then rotate the reference frame back to whatever orientation you desire.

You can't do this with gravity present as it sets a preferred direction.

CB

#### chisigma

##### Well-known member
Air drag proportional to the square of speed only applies at high Reynolds numbers, a balloon will usually be operating at low Reynolds numbers where drag is more nearly proportional to the speed.

CB
That is very interesting!... could You give some more detail, please?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
That is very interesting!... could You give some more detail, please?...

Kind regards

$\chi$ $\sigma$
See the Wikipedia article on Drag.

Having done some calculations and much to my surprise, it would seem that a balloon moving at a speed of the order 0f 0.1 m/s in air is not a low Reynolds number problem, it appears to just squeeze into the high Reynolds number regime (that was for a 1m diameter balloon, a balloon half that diameter will fall out the bottom of the high Reynolds number regime).

Note the wide gap between low (Re<1) and high (Re>1000) Reynolds number regimes.

CB

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#### chisigma

##### Well-known member
See the Wikipedia article on Drag.

Having done some calculations and much to my surprise, it would seem that a balloon moving at a speed of the order 0f 0.1 m/s in air is not a low Reynolds number problem, it appears to just squeeze into the high Reynolds number regime (that was for a 1m diameter balloon, a balloon half that diameter will fall out the bottom of the high Reynolds number regime).

Note the wide gap between low (Re<1) and high (Re>1000) Reynolds number regimes.

CB
Very interesting the Wikipedia article that CB has indicated!... a remarkable fact is that in low Reynolds number regime the drag is linearly dependent from velocity accordingto the formula $\displaystyle\overrightarrow {f}_{d}= -b\ \overrightarrow {v}$ and that means that in such a case. on the basis of what i wrote in a previous post, the problem of motion of a material point subject to gravity and drag can be treated separately along the axes x and y. Along the x-axis the motion equation is...

$\displaystyle v^{\ '}_{x}= - \beta\ v_{x}$ (1)

... where $\beta$ is a constant measured in $s^{-1}$. The solution of (1) is very comfortable...

$\displaystyle v^{\ '}_{x}= - \beta\ v_{x} \implies \frac{d v_{x}}{v_{x}} = - \beta\ dt \implies v_{x}= v_{x}(0)\ e^{- \beta\ t}$ (2)

... and the ordinate x is found integrating (2)...

$\displaystyle x=x_{0} + v_{x}(0)\ \int_{0}^{t} e^{- \beta\ \tau}\ d \tau= x_{0} + \frac{v_{x}(0)}{\beta}\ (1-e^{- \beta\ t})$ (3)

Here we have a first remarkable difference respect to the 'quadratic' case: in that case for any $v_{x}(0) \ne 0$ is $\displaystyle \lim_{t \rightarrow \infty} |x|= \infty$, so that the 'air resistance' isn't an effective 'brake'. In the 'linear' case the limit is finite in any case, so that the 'air resistance' is an effective 'brake'.

But now let's examine what happens along the y-axis, where the motion equation is...

$\displaystyle v^{\ '}_{y}= -g- \beta\ v_{y}$ (4)

Also in this case the solution is comfortable...

$\displaystyle \frac{d v_{y}}{g+\beta\ v_{y}}= - dt \implies \frac{1}{\beta}\ \ln (g+\beta\ v_{y}) = -t + \ln c \implies v_{y}=- \frac{g}{\beta} + \{v_{y}(0)+ \frac{g}{\beta}\}\ e^{- \beta\ t}$ (5)

... and the ordinate y is...

$\displaystyle y=y_{0} - \frac{g}{\beta}\ t + \frac{1}{\beta}\ \{v_{y}(0)+ \frac{g}{\beta}\}\ (1-e^{- \beta\ t})$ (6)

Now I please You all to verify what I wrote and I hope to have time to do some calculations supporting these results...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Shouryya Ray’s ‘unsolved Newton problem’

There is some clarification of what the problem was, and the originality or otherwise of the work on aperiodical >>here<<

CB

#### HallsofIvy

##### Well-known member
MHB Math Helper
Thanks. I had suspected from the start that he had actually solved a special case of the general "unsolved" problem and that appears to be the case.