Up, Up, and Strictly Away

Relevant wiki: Projectile motion - Medium

The answer is independent of the values given.

The shortest distance is attained when the velocity vector $\vec v$ is perpendicular to the segment from the starting point $\vec d$ . It is well-known that $\begin{cases} d_x = v_{x0}t \\ d_y = v_{y0}t - \tfrac12 gt^2 \end{cases}\ \ \ \ \ \ \ \ \ \begin{cases} v_x = v_{x0} \\ v_y = v_{y0} - gt \end{cases}$ They are perpendicular when their dot product is zero, i.e. $0 = d_xv_x + d_yv_y = v_{x0}^2t + v_{y0}^2t - \frac32 v_{y0} gt^2 + \frac12 g^2t^3.$ We wish to find ratios of $v_{y0}$ and $v_{x0}$ for which this equation does not have solutions.

Divide the equation by $t$ ; then $(v_{0x}^2 + v_{0y}^2) - \frac32 v_{0y}(gt) + \frac12 (gt)^2 = C + B(gt) + A(gt)^2 = 0.$ This is a quadratic equation; it has no solutions iff the discriminant is negative: $D = B^2 - 4AC = \left(\frac32v_{0y}\right)^2 - 2v_{0x}^2 - 2v_{0y}^2 < 0.$ $\frac14v_{0y}^2 < 2v_{0x}^2.$ $\tan^2\theta = \frac{v_{0y}^2}{v_{0x}^2} < \frac 2{1/4} = \boxed{8}.$

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To illustrate this, I have plotted the graphs of $d(t)$ for various angles.

The blue graph is for an angle of $65^\circ$ , less than our critical angle. The black graph is for an angle of $75^\circ$ . The red graph shows the angle we calculated, of approx. $70.5^\circ$ . It has a horizontal inflection point toward the end of the motion. If the launch angle is made any greater, this inflection point will bifurcate into a local minimum and a local maximum.

We can also find the location of the projectile at this bifurcation point. If $D = 0$ , the solution of the quadratic above is given by $gt = \frac{-B}{2A} = \frac 32v_{0y}.$ The velocity at that time will be $(v_x,v_y) = (v_{0x},-\tfrac12v_{0y})$ , and the direction of $\vec d$ is perpendicular; so that $\frac{d_y}{d_x} = -\frac{v_x}{v_y} = 2\frac{v_{0x}}{v_{0y}} = \frac2{\tan\theta},$ so that the direction angle of $\vec d$ at that moment is equal to $\phi_d = \text{inv tan} \frac 1{\sqrt 2} \approx 35.3^\circ,$ again independent of the values of $v_0$ and $g$ .

In fact, this direction angle $\phi_d$ is exactly half of the launch angle: $\tan 2\phi_d = \frac{2\tan \phi_d}{1-(\tan \phi_d)^2} = \frac{\sqrt 2}{1 - \tfrac12} = 2\sqrt 2 = \sqrt 8 = \tan\theta.$

Relevant wiki: Problem solving 2D

This doesn't need to be framed as a 3D problem. 2D kinematics works fine. Write down the $x$ and $y$ expressions. I will simply call the downward acceleration $g$ because that is a familiar convention.

$x = V_0 \cos(\theta) t \\ y = V_0 \sin(\theta) t - \frac{1}{2}g t^2$

Distance from launch point (origin):

$D^2 = x^2 + y^2 = V_0^2 \,\cos^2(\theta) t^2 + V_0^2 \, \sin^2(\theta) t^2 - V_0 \, g \, \sin(\theta) t^3 + \frac{1}{4} g^2 t^4 \\ = V_0^2 t^2 - V_0 \, g \, \sin(\theta) t^3 + \frac{1}{4} g^2 t^4$

Rate of change of squared distance must be greater than zero for all values of $t$ :

$\frac{d(D^2)}{dt} = 2 V_0^2 t - 3 V_0 \, g \, \sin(\theta) t^2 + g^2 t^3 > 0 \\ \implies 2 V_0^2 - 3 V_0 \, g \, \sin(\theta) t + g^2 t^2 > 0$

Plugging in numbers gives:

$20000 - 300 \sin(\theta) t + t^2 > 0 \\ c + b \, t + a \, t^2 > 0$

We know that this will be satisfied if there are no real-valued solutions for the quadratic:

$20000 - 300 \sin(\theta) t + t^2 \neq 0$

Determine the critical value of $\theta$ which makes $(b^2 = 4ac)$ :

$90000 \sin^2(\theta) = 80000 \\ \sin^2(\theta) = \frac{8}{9} \\ \tan^2(\theta) = \frac{\sin^2(\theta)}{1 - \sin^2(\theta)} = \frac{\frac{8}{9}}{1 - \frac{8}{9}} = \boxed{8}$

Relevant wiki: Problem solving 2D

The motion is parabolic:

$y = ax^2 + bx$

where $b = \tan\theta$ . Just combine the horizontal equation of motion $x = V_0t\cos\theta$ with the vertical one $y = -g't^2/2 + V_0t\sin\theta$ in order to obtain that.

Now the distance from the origin:

$d^2 = x^2 + y^2 = x^2[1 + (ax + b)^2]$

Derive it with respect to $x$ (which I use here instead of time as independent variable because $x(t)$ is a monotonic function of time):

$2dd' = 2x(2a^2x^2 + 3abx + 1 + b^2)$

The prime denotes derivation with respect to $x$ . Now, $d$ and $x$ are always positive, so $d' > 0$ if the discriminant of the second order polynomial between parenthesis is non-positive:

$\Delta^2 = 9a^2b^2 - 8a^2(1 + b^2) \le 0$

From which we get:

$b^2 \le 8$

Therefore:

$\boxed{\tan^2\alpha = 8}$

The result is already independent of mass, $g'$ and $V_0$ .

Define $\hat{x}$ to be the unit vector parallel to the projection of $\vec{V_{0}}$ onto the plane, and $\hat{y}$ to be the unit vector which points upwards normal to the plane.

Let $\vec{r}(t)=x(t)\hat{x}+y(t)\hat{y}$ be the ball's position vector and let $C$ be the circle whose center is the origin and circumference passes through $\vec{r}(t)$ . The distance between the origin and $\vec{r}(t)$ is increasing if and only if the ball's velocity points outside of the circle. Hence;

$\frac{dy}{dx} \left|_{\vec{r}(t)} > \frac{dy}{dx} \right|_{C}$

Since $\frac{dy}{dx}$ along any circle centered at the origin is $\frac{-x}{y}$ , we get that $\frac{y'}{x'}+\frac{x}{y}>0$ or $yy'+xx'>0$ .

Since this is a kinematic problem, it is true that $y(t)=At+Bt^{2}$ and $x(t)=Ct$ (the only acceleration is in the y-direction)

We get:

$[At+Bt^{2}][A+2Bt]+C^{2}t>0 \quad \Rightarrow \quad (A^{2}+C^{2})+3ABt+2B^{2}t^{2}>0$

Note: we can divide through by $t$ since we only care about when $t>0$ .

For the above to be true for $t>0$ , we must have no positive roots in $t$ . In other words, either the discriminant is negative, or both roots are negative. Based on our knowledge of kinematics, we know that $B$ corresponds with the $y$ acceleration, which is negative. We also know $A$ corresponds with the initial $y$ velocity, which is positive. Hence, $\frac{-3AB}{4B^{2}}$ , the $t$ value which corresponds to the axis of symmetry of the parabola, is positive. Thus, we must have the discriminant be negative.

We get $9A^{2}B^{2}-8B^{2}(A^{2}+C^{2})<0 \quad \Rightarrow \quad \frac{A^{2}}{C^{2}}<8$

We know $A=V_{0}\sin(\theta)$ and $C=V_{0}\cos(\theta)$ , so we get that $tan^{2}(\theta) < 8$ . Since $\tan(\theta)$ is increasing over our interval, we get that $\boxed{\tan^{2}(\alpha)=8}$ satisfies our conditions on $\alpha$ . This is true regardless of our chosen values for $V_{0}$ , $m$ , and $g$ .

Lets call the coordinates of the ball $(\vec{r}(t)=x(t),z(t))$

From the picture we can see that the equation of motion is

$\vec{r}(t) = \left(v_0 \cos(\theta) t, v_0 \sin(\theta)t-\frac{1}{2}g t^2\right)$

Thus the distance of the ball from the axis origin is

$\left|\vec{r}(t)\right|^2 = v^2_0 \cos^2(\theta) t^2+ v^2_0 \sin^2(\theta)t^2-v_0\sin(\theta)g t^3+\frac{1}{4}g^2t^4=At^2+Bt^3+Ct^4$

with $A=v^2_0$ , $B= -v_0\sin(\theta)g$ and $C = \frac{1}{4}g^2$ . We want $\left|\vec{r}(t)\right|$ to be constantly increasing and so the $\left|\vec{r}(t)\right|^2$ to be constantly increasing. That means we demand that

$\frac{d}{dt}(At^2+Bt^3+Ct^4)> 0 \Rightarrow 2At+3Bt^2+4Ct^3>0$

and because time is positive we devide with it

$2A+3Bt+4Ct^2>0$

So for this to be satisfied the distinctive $\Delta=9B^2-8AC$ must be negative! Thus we get

$9v^2_0 \sin^2 (\theta)g^2-8v^2_0g^2<0 \Rightarrow \boxed{\sin^2(\theta)<\frac{8}{9}}$

In the liminiting case where $\theta = \alpha$ , $\sin^2(\alpha)=8/9$ and so

$\tan^2{\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac{8/9}{1/9}=\boxed{8}$

We see from the solution that the result is independent of the gravitational consant g (the planet we do the projectile) the initial velocity $v_0$ and the mass $m$ of the object!

I "cheated": I graphed the distance from the origin in Excel, and varied the angle until I found the extremal value. I stopped refining when it became clear that the answer is 8. See the 3 figures below