2 problems found
A gun is sited on a horizontal plain and can fire shells in any direction and at any elevation at speed \(v\). The gun is a distance \(d\) from a straight railway line which crosses the plain, where \(v^2>gd\). The gunner aims to hit the line, choosing the direction and elevation so as to maximize the time of flight of the shell. Show that the time of flight, \(T\), of the shell satisfies \[ %\frac{2v}{g} \sin \left( \frac12 \arccos \frac{gd}{v^2}\right)\,. g^2 T^2 = 2 v^2 +2 \left(v^4 -g^2d^2\right)^{\frac12}\,. \] Extension: (Not in original paper) Find the time of flight if the gun is constrained so that the angle of elevation \(\alpha \) is not greater than \( 45^\circ\).
Solution: If we fire the gun at an angle to the track, as long as we can travel a horizontal distance \(\geq d\) we can hit the track. Suppose we am at an elevatation \(\alpha\), then \begin{align*} (\uparrow): && s &= ut + \frac12 at^2 \\ && 0 &= v\sin \alpha T - \frac12 g T^2 \\ \Rightarrow &&T &= \frac{2v\sin \alpha}{g}\\ \\ (\rightarrow): && s &= ut \\ && s &= v \cos \alpha T \\ &&&= v\sqrt{1-\sin^2 \alpha} T \\ &&&= vT\sqrt{1 - \frac{g^2T^2}{4v^2}}\\ &&&= \frac{T}{2}\sqrt{4v^2-g^2T^2}\\ \Rightarrow && d & \leq \frac{T}{2}\sqrt{4v^2-g^2T^2} \\ \Rightarrow && 4g^2d^2&\leq g^2T^2(4v^2-g^2T^2) \\ \Rightarrow && 0 &\leq -(g^2T^2)^2 + 4v^2 (g^2T^2)-4g^2d^2 \\ &&&=4v^4-4g^2d^2 -\left (g^2T^2-2v^2 \right)^2 \\ \Rightarrow && \left (g^2T^2-2v^2 \right)^2 & \leq 4v^4-4g^2d^2 \\ \Rightarrow && g^2T^2 &\leq 2v^2+2\sqrt{v^4-g^2d^2} \end{align*} Therefore the maximum value for \(g^2T^2\) is \(2v^2+2\sqrt{v^4-g^2d^2}\) Notice that we are hitting the track directly at \(d\). This is because to maximise the time of flight (for a fixed speed) we want to maximise the angle of elevation. Therefore we want the highest angle where we still hit the track (which is clearly the shortest distance). If we are constraint to \(\alpha \leq 45^\circ\) we know that \(T\) is maximised when \(\alpha = 45^\circ\) (and we will reach the track since the range \(\frac{v^2 \sin 2 \alpha}{g}\) is increasing). Therefore the maximum time is \(T = \frac{\sqrt{2}v}{g}\)
A shell of mass \(m\) is fired at elevation \(\pi/3\) and speed \(v\). Superman, of mass \(2m\), catches the shell at the top of its flight, by gliding up behind it in the same horizontal direction with speed \(3v\). As soon as Superman catches the shell, he instantaneously clasps it in his cloak, and immediately pushes it vertically downwards, without further changing its horizontal component of velocity, but giving it a downward vertical component of velocity of magnitude \(3v/2\). Calculate the total time of flight of the shell in terms of \(v\) and \(g\). Calculate also, to the nearest degree, the angle Superman's flight trajectory initially makes with the horizontal after releasing the shell, as he soars upwards like a bird. {[}Superman and the shell may be regarded as particles.{]}
Solution: The particle has initial velocity \(\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}\) and acceleration \(\displaystyle \binom{0}{-g}\). It will have zero vertical speed (ie be at the top of its trajectory) when \(t = \frac{\sqrt{3}v}{2g}\). Since \(0 = v^2-u^2 + 2as\) the height achieved will be \(\frac{3v^2}{8g}\) At this point it will need to travel the same distance again, but this time the initial speed is \(\frac{3v}{2}\) so: \begin{align*} && \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\ \Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\ \Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g} \end{align*} Therefore the total time is: \begin{align*} \l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g} \end{align*} \begin{align*} COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\ \Rightarrow &&v_y &= \frac34 v \\ COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\ \Rightarrow && V &= \frac{13}6\\ \end{align*} Therefore superman is now travelling at a vector of \(\displaystyle \binom{\frac{13}6}{\frac34}v\) ie an angle of \(\tan^{-1} \frac 9{26}\) to the horizontal, approximately \(19^\circ\)