User chrisjb - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:16:20Z http://mathoverflow.net/feeds/user/29287 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33807/angle-maximizing-the-distance-of-a-projectile/114063#114063 Answer by ChrisJB for Angle Maximizing the Distance of a Projectile ChrisJB 2012-11-21T14:50:15Z 2012-11-28T01:34:36Z <p>Consider what path is traced out by the projectile in the 2d velocity space (horizontal velocity x-axis; horizontal is "after rotating up so the ground is flat, gravity no longer vertical"). It starts somewhere on a circular arc, and thereafter follows a path 'down and to the right' at an angle $\phi$ to the vertical, at constant speed (corresponding to strength of g). Trace the line until it reaches the horizontal. This forms a triangle along with the origin.</p> <p>The total distance travelled is just (total time in air)x(speed at max height), which is just proportional to the area of the triangle for fixed $\phi$. Possible initial angles give a family of triangles, with one side of fixed length and the opposite angle also fixed ($\pi/2 - \phi$); thus they fit in a circle and the maximum is clearly when the triangle is isosceles.</p> <p><strong>Added 11/24/2012</strong> </p> <p>Here is a slightly more detailed solution along with a picture. This was written up by Davidac897 with the aid of Barry Cipra's writeup of ChrisJB's solution.</p> <p><img src="https://web.math.princeton.edu/~dcorwin/mogeopic.png" alt="alt text"></p> <p>If you rotate the system so the ground is flat, you'll be firing at angle $\theta' = \theta + \phi$ into a medium where gravity points down and to the right at angle $\phi$. Therefore</p> <p>In the velocity plane, the trajectory starts at $P$ and follows a straight line at angle $\pi/2 - \phi$ to the $v_x$ axis, through a point $Q$ on the $v_x$ axis, down to a point $P'$ with $v_y$ coordinate the negative of that of $P$ (this is true if we fire in a non-rotated frame, and the only difference here is that there is an extra force component in the $x$-direction).</p> <p>In particular $\angle QOP = \theta' = \theta+\phi$, $\angle PQO = \frac{\pi}{2}-\phi$, and the segments intersect at $R$ in a right angle, where $\overline{QR}$ is the horizontal in the non-rotated frame. Then $OQ$ is the average horizontal speed of the projectile, and $OA$ is proportional to the total time in the area because it is half the total change in vertical velocity. Therefore, the area of the triangle, which is $\frac{(OQ)(OA)}{2}$, is proportional to the total distance traveled.</p> <p>As remarked above, $OP$ is fixed, as is $\angle PQO$, and the area is maximized when $Q$ is the apex of an isosceles triangle, so $2 \angle QOP + \angle PQO = 2\theta + 2 \phi + \frac{\pi}{2}-\phi = \pi$, or $$\theta = \frac{\pi}{4}-\frac{\phi}{2}.$$</p>