Subfield: Mechanics

Problem T1. Focus on sketches

Part A. Ballistics

A ball, thrown with an initial speed $v_{0}$, moves in a homogeneous gravitational field in the $x-z$ plane, where the $x$-axis is horizontal, and the $z$-axis is vertical and antiparallel to the free fall acceleration $g$. Neglect the effect of air drag.
Context question:
i. By adjusting the launching angle for a ball thrown with a fixed initial speed $v_{0}$ from the origin, targets can be hit within the region given by

$$
z \leq z_{0}-k x^{2}
$$

You can use this fact without proving it. Find the constants $z_{0}$ and $k$.
Context answer:
\boxed{$\frac{v_{0}^{2}}{ 2 g}$ , $\frac{g}{2 v_{0}^{2}}$}


Context question:
ii. The launching point can now be freely selected on the ground level $z=0$, and the launching angle can be adjusted as needed. The aim is to hit the topmost point of a spherical building of radius $R$ (see fig.) with the

<img_4494>
minimal initial speed $v_{0}$. Bouncing off the roof prior to hitting the target is not allowed. Sketch qualitatively the shape of the optimal trajectory of the ball (use the designated box on the answer sheet). Note that the marks are given only for the sketch.
Context answer:
<img_4473>


Question: iii. What is the minimal launching speed $v_{\min }$ needed to hit the topmost point of a spherical building of radius $R$ ?

<img_4455>

La Geode, Parc de la Villette, Paris. Photo: katchooo/flickr.com

Solution:
The brute force approach would be writing down the condition that the optimal trajectory intersects with the building at two points and touches at one. This would be described by a fourth order algebraic equation and therefore, it is not realistic to accomplish such a solution within a reasonable time frame.

Note that the interior of the building needs to lie inside the region where the targets can be hit with a stone thrown from the top with initial speed $v_{\min }$. Indeed, if we can throw over the building, we can hit anything inside by lowering the throwing angle. On the other hand, the boundary of the targetable region needs to touch the building. Indeed, if there were a gap, it would be possible to hit a target just above the point where the optimal trajectory touches the building; the trajectory through that target wouldn't touch the building anywhere, hence we arrive at a contradiction.

So, with $v_{0}$ corresponding to the optimal trajectory, the targetable region touches the building; due to symmetry, overall there are two touching points (for smaller speeds, there would be four, and for larger speeds, there would be none). With the origin at the top of the building, the intersection points are defined by the following system of equations:

$$
x^{2}+z^{2}+2 z R=0, \quad z=\frac{v_{0}^{2}}{2 g}-\frac{g x^{2}}{2 v_{0}^{2}}
$$

Upon eliminating $z$, this becomes a biquadratic equation for $x$ :

$$
x^{4}\left(\frac{g}{2 v_{0}^{2}}\right)^{2}+x^{2}\left(\frac{1}{2}-\frac{g R}{v_{0}^{2}}\right)+\left(\frac{v_{0}^{2}}{4 g}+R\right) \frac{v_{0}^{2}}{g}=0
$$

Hence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes:

$$
\left(\frac{1}{2}-\frac{g R}{v_{0}^{2}}\right)^{2}=\frac{1}{4}+\frac{g R}{v_{0}^{2}} \Longrightarrow \frac{g R}{v_{0}^{2}}=2
$$

Bearing in mind that due to the energy conservation law, at the ground level the squared speed is increased by $4 g R$. Thus we finally obtain

$$
v_{\min }=\sqrt{v_{0}^{2}+4 g R}=3 \sqrt{\frac{g R}{2}} .
$$

Final answer: $v_{\min }=3 \sqrt{\frac{g R}{2}}$