3 problems found
The point \(O\) is at the top of a vertical tower of height \(h\) which stands in the middle of a large horizontal plain. A projectile \(P\) is fired from \(O\) at a fixed speed \(u\) and at an angle \(\alpha\) above the horizontal. Show that the distance \(x\) from the base of the tower when \(P\) hits the plain satisfies \[ \frac{gx^2}{u^2} = h(1+\cos 2\alpha) + x \sin 2\alpha \,. \] Show that the greatest value of \(x\) as \(\alpha\) varies occurs when \(x=h\tan2\alpha\) and find the corresponding value of \(\cos 2\alpha\) in terms of \(g\), \(h\) and \(u\). Show further that the greatest achievable distance between \(O\) and the landing point is \(\dfrac {u^2}g +h\,\).
Solution: \begin{align*} \rightarrow: && x &= u \cos \alpha t\\ \Rightarrow && t &= \frac{x}{u \cos \alpha}\\ \uparrow: && -h &= u\sin \alpha t- \frac12gt^2 \\ && - h &= x\tan \alpha - \frac12 g \frac{x^2}{u^2}\sec^2 \alpha \\ \Rightarrow && \frac{gx^2}{u^2} &= h(2\cos^2 \alpha) + x2 \tan \alpha \cos^2 \alpha \\ &&&= h(1 + \cos 2 \alpha) + x \sin 2\alpha \\ \frac{\d}{\d \alpha}: && \frac{g}{u^2} 2 x \frac{\d x}{\d \alpha} &= -2h \sin 2 \alpha + 2x \cos 2 \alpha +\frac{\d x}{\d \alpha} \sin 2 \alpha \\ \Rightarrow && \frac{\d x}{\d \alpha} \left ( \frac{2xg}{u^2} - \sin 2 \alpha \right) &= 2\cos 2 \alpha (x -h \tan 2 \alpha) \end{align*} Since the turning point will be a maximum must be \(x = h \tan 2 \alpha\). Therefore, let \(c = \cos 2 \alpha\) \begin{align*} && \frac{gh^2}{u^2} \tan^2 2 \alpha &= h(1 + \cos 2 \alpha) + h \tan 2 \alpha \sin 2 \alpha \\ \Rightarrow && \frac{gh}{u^2}(c^{-2}-1) &= 1+c+\frac{1-c^2}{c} \\ \Rightarrow && \frac{gh(1-c^2)}{u^2c^2} &= \frac{c+c^2+1-c^2}{c}\\ &&&= \frac{1+c}{c} \\ \Rightarrow && \frac{gh(1-c)}{u^2c} &= 1 \\ \Rightarrow && u^2c &= gh(1-c) \\ \Rightarrow && c(u^2+gh) &= gh \\ \Rightarrow && \cos 2 \alpha &= \frac{gh}{u^2+gh} \\ \\ \Rightarrow && d_{max}^2 &= h^2 + h^2 \tan^2 2 \alpha \\ &&&= h^2\sec^2 2 \alpha \\ &&&= h^2 \frac{(u^2+gh)^2}{g^2h^2} \\ &&&= \frac{(u^2+gh)^2}{g^2} \\ &&&= \left (\frac{u^2}{g}+h \right)^2 \\ \Rightarrow && d_{max} &= \frac{u^2}{g}+h \end{align*}
A child is playing with a toy cannon on the floor of a long railway carriage. The carriage is moving horizontally in a northerly direction with acceleration \(a\). The child points the cannon southward at an angle \(\theta\) to the horizontal and fires a toy shell which leaves the cannon at speed \(V\). Find, in terms of \(a\) and \(g\), the value of \(\tan 2\theta\) for which the cannon has maximum range (in the carriage). If \(a\) is small compared with \(g\), show that the value of \(\theta\) which gives the maximum range is approximately \[ \frac \pi 4 + \frac a {2g}, \] and show that the maximum range is approximately \(\displaystyle \frac {V^2} g + \frac {V^2a}{g^2}. \)
Solution: \begin{align*} && s_x &= V \cos \theta t + \frac12at^2 \\ && s_y &= V \sin \theta t - \frac12 gt^2 \\ \Rightarrow && T &= \frac{2V \sin \theta}g \\ \Rightarrow && s_{max} &= \frac{2V^2 \sin \theta \cos \theta}{g} + \frac12a \frac{4V^2 \sin^2 \theta}{g^2} \\ &&&= (g \sin 2 \theta+2a\sin^2 \theta)\frac{V^2}{g^2} \\ && \frac{\d s_{max}}{\d \theta} &= (2g \cos 2 \theta +4 a \cos \theta \sin \theta)\frac{V^2}{g^2} \\ &&&= (2g \cos 2\theta + 2a \sin2 \theta) \frac{V^2}{g^2} \\ \Rightarrow && \tan 2\theta &= -\frac{a}{g} \\ \Rightarrow && 2 \theta &\in (\frac{\pi}2, \pi) \\ \Rightarrow && \tan \left (\frac{\pi}{2} - 2 \theta\right) &=-\frac{a}{g} \\ \Rightarrow && \frac{\pi}{2} - 2 \theta&\approx -\frac{a}{g} \\ \Rightarrow && \theta &\approx \frac{\pi}{4} + \frac{a}{2g} \\ \\ && s_{max} & \approx \left (g \sin \left (\frac{\pi}{2} + \frac{a}{g} \right)+2a\sin^2 \left ( \frac{\pi}{4} + \frac{a}{2g}\right)\right)\frac{V^2}{g^2} \\ &&&\approx \left (g \cdot 1+2a\left( \frac{1}{\sqrt{2}}(\frac{a}{2g}+1)\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\left(1+\frac{a}{g}\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\right)\frac{V^2}{g^2} \\ &&&= \frac{V^2}{g} + \frac{V^2a}{g} \end{align*}
A cannon-ball is fired from a cannon at an initial speed \(u\). After time \(t\) it has reached height \(h\) and is at a distance \(\sqrt{x^{2}+h^{2}}\) from the cannon. Ignoring air resistance, show that \[ \tfrac{1}{4}g^{2}t^{4}-(u^{2}-gh)t^{2}+h^{2}+x^{2}=0. \] Hence show that if \(u^{2}>2gh\) then the horizontal range for a given height \(h\) and initial speed \(u\) is less than or equal to \[ \frac{u\sqrt{u^{2}-2gh}}{g}. \] Show that there is always an angle of firing for which this value is attained.
Solution: Suppose it is fired with angle to the horizontal \(\alpha\), then \begin{align*} \rightarrow: && x &= u\cos \alpha \cdot t \\ \uparrow: && h &= u \sin \alpha \cdot t - \frac12 g t^2 \\ \Rightarrow && u\cos \alpha &= \frac{x}{t} \\ && u \sin \alpha &= \frac{h + \frac12 gt^2}{t} \\ \Rightarrow && u^2 &= \frac{x^2}{t^2} + \frac{(h + \frac12 gt^2)^2}{t^2} \\ \Rightarrow && 0 &= x^2+h^2-u^2t^2+ght^2+\tfrac14 g^2 t^4 \\ &&&= \tfrac14 g^2 t^4 - (u^2 - gh)t^2 + h^2 + x^2 \end{align*} For a distance \(x\) to be achievable there must be a root to this quadratic in \(t^2\), ie \begin{align*} && 0 &\leq \Delta = (u^2-gh)^2 - 4 \cdot \tfrac14 g^2 (h^2 + x^2) \\ \Rightarrow && x^2 &\leq \frac{(u^2-gh)^2}{g^2} - h^2 \\ &&&= \frac{u^4+g^2h^2 - 2ghu^2-g^2h^2}{g^2} \\ &&&= \frac{u^2(u^2-2gh)}{g^2} \\ \Rightarrow && x &\leq \frac{u\sqrt{u^2-2gh}}{g} \end{align*} This is achieved when \begin{align*} && t^2 &= \frac{u^2-gh}{\tfrac12g^2}\\ &&&= \frac{2(u^2-gh)}{g^2} \\ \Rightarrow && \cos \alpha &= \frac{u\sqrt{u^2-2gh}}{g} \cdot \frac{g}{\sqrt{2(u^2-gh)}} \frac{1}{u} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} ie when \(\alpha = \frac{\pi}{4}\)