Always going further

We know the particle's trajectory is a parabola, and it goes through $P$ (which we can define to be the origin). The key idea is that neither the particle's initial speed nor the magnitude of the gravitational acceleration have any effect on the desired property of always moving away from the origin.

This means we can greatly simplify the algebra by letting the equation of the trajectory be $y=ax-x^2$ . The gradient of this curve at $O$ is the tangent of the initial angle of elevation - ie, $a=\tan{\theta}$ .

We want the distance of the particle from $O$ (or equivalently its square) to be increasing for positive $x$ . So define

$f(x) = y^2+x^2 = x^4 - 2ax^3 + (a^2+1)x^2$

Then $f'(x) = 2x(2x^2 - 3ax + a^2+1)$

For the distance to be increasing, we need this to be positive for $x>0$ , ie $2x^2 - 3ax + a^2+1 = 0$ must have no positive roots. But, by inspection, if this expression has real roots, they can only be positive; so we need it to have negative discriminant. The critical case (discriminant zero) is therefore when

$9a^2-4\cdot2(a^2+1) = a^2 - 8 = 0$

This gives $a = \sqrt8$ and $\theta=\boxed{70.5287\ldots}$

Give the parabola a trajectory $y=ax-x^{2}$ which launches from the origin at an angle such that $\tan \theta = a$

We seek a value of $a$ such that no perpendicular to the parabola will pass above the origin after the object reaches its peak.

At any point $(p,ap-p^{2})$ the parabola has slope $a-2p$ and the perpendicular has slope $\frac{1}{2p-a}$ . So the line throught the point has equation $y=\frac{1}{2p-a}(x-p)+ap-p^{2}$ . For this to go through the origin, $x=0$ and $y=0$ , thus

$0=\frac{p}{a-2p}+ap-p^{2}$

$0=-2p^{2}+3ap-a^{2}-1$

If the discriminant of this quadratic in $p$ is zero, we will keep the perpendicular from rising above zero.

$(3a)^{2}-4(-2)(-a^{2}-1)=0$

$a=\sqrt{8}$ This is the slope.

Thus the angle we seek is $\tan^{-1}\sqrt{8} \approx \boxed{70.5}^{\circ}$

We start by considering the components of the initial velocity. The magnitude of the initial velocity does not actually matter but the ratio of the components does. Let the initial velocity in the y-direction be $v_{y0}$ and in the x-direction $v_{x0}$ . We can now write the velocities as functions of time: $v_y (t)=v_{y0}-gt$ and $v_x (t) = v_{x0}$ . The x-component of the velocity is constant throughout its journey but the y-component is influenced by gravity. Next we write down the position functions: $y(t)=v_{y0} t - \frac{1}{2} gt^2$ and $x(t)=v_{x0}t$ . The distance between the object and the point P, $r(t)$ , can now be found using pythagoras. The idea is that we find a relationship between $v_{y0}$ and $v_{x0}$ , such that the time-derivative of $r(t)$ is always positive. $r(t) = \sqrt((v_{x0}t)^2 + (v_{y0}t-\frac{1}{2}gt^2))$ . However, if $r(t)$ is always increasing then so is $(r(t))^2$ , so we can just consider the derivative of this for simplicity. Therefore, the equation we set up is this: $\frac{d \, (r(t))^2}{dt} \, > \, 0$ for all positive values of $t$ . The only left now is to do the math.

$(r(t))^2 =(v_{x0}t)^2 + (v_{y0}t-\frac{1}{2}gt^2)^2$

$\frac{d \, (r(t))^2}{dt} = 2v_{x0}^2t +2(v_{y0}t-\frac{1}{2}gt^2)(v_{y0}-gt)$ > 0

$2v_{x0}^2t + 2v_{y0}^2 t - 3v_{y0} g t^2 + g^2t^3 > 0$

$t(2v_{x0}^2 + 2v_{y0}^2 - 3v_{y0} g t + g^2t^2) > 0$ where $t>0$

$2v_{x0}^2 + 2v_{y0}^2 - 3v_{y0} g t + g^2t^2 > 0$

Now this equation has to be true for all values of $t$ , so if we treat it as a quadratic with $t$ as the variable then the discriminant must be negative. Only then will the rate of change of the distance between P and the object always be positive. Therefore, we are left with the equation below.

$( - 3v_{y0} g)^2 - 4(g^2)(2v_{x0}^2 + 2v_{y0}^2) < 0$

$9v_{y0}^2g^2 - 8g^2v_{x0}^2 - 8g^2v_{y0}^2 < 0$

$9v_{y0}^2 - 8v_{x0}^2 - 8v_{y0}^2 < 0$

$(\frac{v_{y0}}{v_{x0}})^2 - 8 < 0$

$\theta$ is the angle between the initial direction of the object and the horizontal. So if we imagine a triangle with $\theta$ as one of the angles then $v_{y0}$ will be the length of the opposing side and $v_{x0}$ the adjacent. Therefore, we can write that: $\tan \theta =\frac{v_{y0}}{v_{x0}}$ .

$(\tan \theta)^2 < 8$

$\tan \theta < 2\sqrt{2}$

$\theta < \arctan(2\sqrt{2})$

So the maximum angle will be: $\theta = 70.5$ degrees.

To assure that I set up the equations correctly, I started a bit further back: I solved the differential equation of the vertical motion. I recognized that the acceleration of gravity and the initial upward velocity factored out of the equations. Therefore, I set those values to 1. I also assumed, as is usually the case, that the gravitational field was not diverging and not decreasing with height and since this is the classical mechanics category that relativistic effects were ignored.

$y''(t)=g,y'(0)=\sin (\theta ),y(0)=0 \Longrightarrow \frac{1}{2} \left(2 t \sin (\theta )-t^2\right)$ and $t \cos (\theta )$

$\frac{\partial \sqrt{x(t)^2+y(t)^2}}{\partial t} \Longrightarrow \frac{\frac{1}{2} (2 \sin (\theta )-2 t) \left(2 t \sin (\theta )-t^2\right)+2 t \cos ^2(\theta )}{2 \sqrt{\frac{1}{4} \left(2 t \sin (\theta )-t^2\right)^2+t^2 \cos ^2(\theta )}}$

Solving for $\theta$ :

$\frac{\frac{1}{2} (2 \sin (\theta )-2 t) \left(2 t \sin (\theta )-t^2\right)+2 t \cos ^2(\theta )}{2 \sqrt{\frac{1}{4} \left(2 t \sin (\theta )-t^2\right)^2+t^2 \cos ^2(\theta )}}=0\land 1\leq t\leq 2 \Longrightarrow \\ \theta=2 \tan ^{-1}\left(\frac{-\sqrt{\frac{-t^4+5 t^2-4}{\left(t^2+2\right)^2}} t^2-2 \sqrt{\frac{-t^4+5 t^2-4}{\left(t^2+2\right)^2}}+3 t}{t^2+2}\right)\lor \\ \theta =2 \tan ^{-1}\left(\frac{\sqrt{\frac{-t^4+5 t^2-4}{\left(t^2+2\right)^2}} t^2+2 \sqrt{\frac{-t^4+5 t^2-4}{\left(t^2+2\right)^2}}+3 t}{t^2+2}\right)$

Taking the derivative of those functions of $t$ for $\theta$ and solving for the derivative being $0$ gave a time of $t=\sqrt{2}$ . Using that value of $t$ , in the previous equations give the result for $0\leq\theta\leq\frac{\pi}{2}$ of $2 \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \approx 70.5287793655093^{\circ}$ .

The area in $\theta$ and $t$ parameter space where the effect occurs is the area above the curve toi the following plot:

Suppose the projectile starts off with a velocity with horizontal component $v_x$ and vertical component $v_y$ . The position then is described by

$x(t)=v_xt$

$y(t)=v_yt-\frac{1}{2}gt^2$ .

The distance squared is $d^2=x^2+y^2=v_x^2t^2+v_y^2t^2-v_ygt^3+\frac{1}{4}g^2t^4$ .

If the distance reaches a maximum at a certain point, the derivative of this must be 0 at some $t>0$ :

$2v_x^2t+2v_y^2t-3v_ygt^2+g^2t^3=0$

Dividing by t gives $g^2t^2 - (3v_yg)t+2v_x^2+2v_y^2=0$ , which is a quadratic polynomial having real solutions when the discriminant $b^2-4ac$ is nonnegative. Setting it to 0 gives the border case where there is just a solution: $9v_y^2g^2=8g^2v_x^2+8g^2v_y^2$ .

This solves as $v_y^2=8v_x^2$ so that $\theta = \arctan(\frac{v_y}{v_x})= \arctan(\sqrt{8}) =\boxed{70.5°}$

My wife is watching Big Bang Theory, and as I was typing arctan they discussed the inverse tangent function. I guess that means I must be a nerd...

Method 1: $\dfrac{\mathrm{d}r}{\mathrm{d}t}>0$

Write the parametric equations of motion of the particle with respect to time: $x(t)=v_0t\cos\theta,\:\:\:\:\:\:\:\:y(t)=v_0t\sin\theta-\dfrac{1}{2}gt^2$

Now, let $\vec{r}$ be the position vector of the particle. As it is always going away from the starting point, $r$ must always increase in magnitude, so $\dfrac{\mathrm{d}r}{\mathrm{d}t}>0$ must always hold. It is easy to obtain the expression of $r(t)$ using the Pythagorean Theorem: $r^2(t)=x^2(t)+y^2(t)\implies r(t)=\sqrt{v_0^2t^2-v_0gt^3\sin\theta+\dfrac{1}{4}gt^4}$

To evaluate the condition $\dfrac{\mathrm{d}r}{\mathrm{d}t}>0$ , it is sufficient to evaluate $\dfrac{\mathrm{d}r^2}{\mathrm{d}t}>0$ thanks to the power rule for functions. This is equivalent to: $2v_0^2t-3v_0gt^2\sin\theta+g^2t^3>0$ .

Rearranging gives: $\sin\theta<\dfrac{2v_0^2t+g^2t^3}{3v_0gt^2}\equiv f(t)$

The easiest thing to do here, in my opinion, is to find the minimum of $f(t)$ w.r.t. time: $\dfrac{\mathrm{d}f}{\mathrm{d}t}=0\implies t=\dfrac{v_0\sqrt{2}}{g}$ , which gives $\sin\theta<\frac{2\sqrt{2}}{3}$ . $\sin(x)$ is strictly increasing on $\left[0,\dfrac{\pi}{2}\right]$ and therefore $\theta<\arcsin\left(\frac{2\sqrt{2}}{3}\right)$ and therefore $\boxed{\theta < 70.528^\circ}$ .

Non-derivative solution making use of the dot-product

The radial velocity, $v_r=\dfrac{\mathrm{d}r}{\mathrm{d}t}$ can also be written in dot-product form as $\vec{r}\cdot\vec{v}$ , where $\vec{v}$ is the instantaneous velocity. Now, one has to write $\vec{r}(t)$ and $\vec{v}(t)$ in terms of $\vec{v_0},t,\vec{g}$ as follows: $\vec{r}(t)=\vec{v_0}t+\frac{\vec{g}}{2}t^2,\:\:\:\:\:\:\:\: \vec{v}(t)=\vec{v_0}+\vec{g}t$ And therefore, evaluating the dot product and dividing both sides by $\dfrac{v_0t^2g\sqrt{2}}{2}\neq 0$ one finds that: $\dfrac{v_0\sqrt{2}}{gt}+\dfrac{gt}{v_0\sqrt{2}}>\dfrac{3\sin\theta\sqrt{2}}{2}$ Rearranging gives: $\sin\theta<\dfrac{\sqrt{2}}{3}\left(\dfrac{v_0\sqrt{2}}{gt}+\dfrac{gt}{v_0\sqrt{2}}\right)$ Notice that $\dfrac{v_0\sqrt{2}}{gt}=\left(\dfrac{gt}{v_0\sqrt{2}}\right)^{-1}$ , and therefore one can use the famous inequality $x+x^{-1}\ge 2$ , with the equality case only holding for $x=1\iff v_0\sqrt{2}=gt$ (identical to the $t$ that minimises $f(t)$ in the previous solution). This gives $\sin\theta<\dfrac{2\sqrt{2}}{3}$ so again, $\boxed{\theta < 70.528^\circ}$ .

I have tried to keep my solution tied to the physical situation. I like the way both the initial velocity and the strength of gravity cancel out in the calculation.

Launch the projectile from the origin at an angle $\theta$ to the horizontal x-axis. Resolve the velocity into a uniform motion in the x-direction and a downwardly accelerated motion. The coordinates of the particle are then parametrised by t, the time of flight:-

$x=Vt\cos\theta \\ y=Vt\sin\theta -\frac{gt^2}{2}$

The square of the distance from the origin now comes by Pythagoras

$s^2 = x^2 +y^2 =V^2t^2 +\frac{g^2t^4}{4}-gVt^3\sin\theta$

If the distance is always increasing the time derivative of this expression must be greater than zero:

$2V^2 t + g^2 t^3 - 3gVt^2\sin\theta>0$

Solving this for $\sin\theta$ gives

$\sin\theta>\frac{2V^2t+g^2t^3}{3Vgt^2}$

Now, how small can this right hand expression be? Rewrite it as

$\sin\theta>\frac{\sqrt{2}}{3}(\frac{\sqrt{2}V}{gt}+\frac{gt}{\sqrt{2}V})$

The term in brackets is of the form $x+\frac{1}{x}$ which has a minimum value of 2, so

$\sin\theta>\frac{2\sqrt{2}}{3}$

and so

$\theta>70.5$ degrees

At any time t, the angle of velocity from horizontal is €, and angle of position vector of particle is ¢ from horizontal. We just make sure that : tan(¢)>tan(€)