Consider any projectile O and its path. Assume that drag or any other type of non-conservative or resisting force doesn't exist and that only that gravitational pull of the earth exists. Note: Let the acceleration due to gravity g remain constant.

Hence, the trajectory of this projectile near the earth's surface will be a parabola. Take any point A on this parabola, and drop a perpendicular to the parabola's directrix at B. With A as the center and AB as the radius draw a circle which will surely meet the focus.

Let this circle cross the vertical line of symmetry of the parabola(i.e. the line of symmetry is a secant to the circle). Is this possible and if yes, what happens in this physical situation? enter image description here

For Your Information ---

  • The distance from the focus to the launch point of a projectile is $v^2/(2g)$
  • This distance is always true regardless of the angle of projection.
  • The x coordinate of the vertex of a parabola and its focus are equal.
  • The distance from any point to the focus is equal to the perpendicular distance from that point to the directrix.(Eg: AF = AB)
  • The initial velocity of water coming out the cylindrical vessel is given by.. $$v = \sqrt{2g(H-h)}$$

Here is an example problem that illustrates such a case. Give it a try if you want to.

I came across this problem statement, in this book... https://www.cambridge.org/core/books/200-more-puzzling-physics-problems/B4DC45F5A40D4B792653B407975F3A41

At the top of a long incline that makes an angle θ with the horizontal, there is a cylindrical vessel containing water to a depth H. A hole is to be drilled in the wall of the cylinder, so as to produce a water jet that lands a distance d down the incline. How far, h, from the bottom of the vessel should the hole be drilled in order to make d as large as possible? What is this maximum value of d?

enter image description here

Note:F is the focus of the parabolic path taken by the projectile and e is the directrix.

In the solution for this problem I found the following statements...

Consider an arbitrary point P on the incline, and denote the point on the directrix directly above it by Q. If there is at least one possible water jet trajectory passing through P, then its focus is as far from P as P is from the directrix e. Now consider a circle, denoted by k, that has center P and radius PQ. Let P approach the vessel on the inclined plane. At some particular point P', the associated circle k', with center P' and radius P'Q', touches the y-axis. The point at which it touches gives the focus F of the parabolic path followed by water that falls the maximum distance down the slope. enter image description here

If the incline is steep enough, it can happen that the distance OF is larger than H (corresponding to a launch point below ground). By investigating the (physically) limiting case of OF = H, we can obtain the critical angle θ' for the slope of the incline.

If θ > θ', the best that can be done is to drill the hole at the bottom of the vessel (at O, where h = 0). Then, circle k' no longer touches the y-axis, but crosses it at point O.


1 Answer 1


Without loss of generality, let $x = v_0 t$ and $y = -\frac{1}{2}g t^2$ so $y = -g x^2/(2v_0^2).$ As you have found, the focus is at $(0,-v_0^2/(2g))$ and the directrix is the line $y=v_0^2/(2g)$. The circle through a point on the parabola and the focus is given by $(x-x_0)^2+(y-y_0)^2=r^2$ where $y_0 = -gx_0^2/(2v_0^2)$ and $r=(v_0^2/g+gx_0^2/v_0^2)/2$. Setting $x=0$ and solving for $y$ we find $y_1 = -v_0^2/(2g)$ and $y_2= v_0^2/(2g)-gx_0^2/v_0^2$. Then $y_1=y_2$ if and only if $x_0 = \pm v_0^2/g$. Thus, are only two points on the parabola for which there is one point of intersection with the line of symmetry. This occurs when the particle is twice the focal length from the line of symmetry. For all other points the circle intersects the line of symmetry at two points. Thus, this occurs at almost all points on the trajectory---these points are not special.

enter image description here

Figure 1. Trajectory for $v_0=10\,\mathrm{m/s}$, $g=10\,\mathrm{m/s^2}$.

  • $\begingroup$ Well explained. Thanks! $\endgroup$
    – Swams2004
    May 27, 2019 at 10:25
  • $\begingroup$ @Swams2004: I am glad to help! $\endgroup$
    – user26872
    May 27, 2019 at 12:38

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