Problems

Point \(A\) is a distance \(h\) above ground level and point \(N\) is directly below \(A\) at ground level. Point \(B\) is also at ground level, a distance \(d\) horizontally from \(N\). The angle of elevation of \(A\) from \(B\) is \(\beta\). A particle is projected horizontally from \(A\), with initial speed \(V\). A second particle is projected from \(B\) with speed \(U\) at an acute angle \(\theta\) above the horizontal. The horizontal components of the velocities of the two particles are in opposite directions. The two particles are projected simultaneously, in the vertical plane through \(A\), \(N\) and \(B\). Given that the two particles collide, show that \[d\sin\theta - h\cos\theta = \frac{Vh}{U}\] and also that

1. \(\theta > \beta\);
2. \(U\sin\theta \geqslant \sqrt{\dfrac{gh}{2}}\);
3. \(\dfrac{U}{V} > \sin\beta\).

1. Two particles move on a smooth horizontal surface. The positions, in Cartesian coordinates, of the particles at time \(t\) are \((a+ut\cos\alpha \,,\, ut\sin\alpha)\) and \((vt\cos\beta\,,\, b+vt\sin\beta )\), where \(a\), \(b\), \(u\) and \(v\) are positive constants, \(\alpha\) and \(\beta\) are constant acute angles, and \(t\ge0\). Given that the two particles collide, show that \[ u \sin(\theta+\alpha) = v\sin(\theta +\beta)\,, \] where \(\theta \) is the acute angle satisfying \(\tan\theta = \dfrac b a\).
2. A gun is placed on the top of a vertical tower of height \(b\) which stands on horizontal ground. The gun fires a bullet with speed \(v\) and (acute) angle of elevation \(\beta\). Simultaneously, a target is projected from a point on the ground a horizontal distance \(a\) from the foot of the tower. The target is projected with speed \(u\) and (acute) angle of elevation \(\alpha\), in a direction directly away from the tower. Given that the target is hit before it reaches the ground, show that \[ 2u\sin\alpha (u\sin\alpha - v\sin\beta) > bg\,. \] Explain, with reference to part (i), why the target can only be hit if \(\alpha > \beta\).

A particle is projected at an angle of elevation \(\alpha\) (where \(\alpha>0\)) from a point \(A\) on horizontal ground. At a general point in its trajectory the angle of elevation of the particle from \(A\) is \(\theta\) and its direction of motion is at an angle \(\phi\) above the horizontal (with \(\phi\ge0\) for the first half of the trajectory and \(\phi\le0\) for the second half). Let \(B\) denote the point on the trajectory at which \(\theta = \frac12 \alpha\) and let \(C\) denote the point on the trajectory at which \(\phi = -\frac12\alpha\).

1. Show that, at a general point on the trajectory, \(2\tan\theta = \tan \alpha + \tan\phi\,\).
2. Show that, if \(B\) and \(C\) are the same point, then \( \alpha = 60^\circ\,\).
3. Given that \(\alpha < 60^\circ\,\), determine whether the particle reaches the point \(B\) first or the point \(C\) first.

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

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Solution:

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\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

A particle is projected under gravity from a point \(P\) and passes through a point \(Q\). The angles of the trajectory with the positive horizontal direction at \(P\) and at \(Q\) are \(\theta\) and \(\phi\), respectively. The angle of elevation of \(Q\) from \(P\) is \(\alpha\).

1. Show that \(\tan\theta +\tan\phi = 2\tan\alpha\).
2. It is given that there is a second trajectory from \(P\) to \(Q\) with the same speed of projection. The angles of this trajectory with the positive horizontal direction at \(P\) and at \(Q\) are \(\theta'\) and \(\phi'\), respectively. By considering a quadratic equation satisfied by \(\tan\theta\), show that \(\tan(\theta+\theta') = -\cot\alpha\). Show also that \(\theta+\theta'=\pi+\phi+\phi'\,\).

A particle \(P_1\) is projected with speed \(V\) at an angle of elevation \({\alpha}\,\,\,( > 45^{\circ})\,,\,\,\,\) from a point in a horizontal plane. Find \(T_1\), the flight time of \(P_1\), in terms of \({\alpha}, V \hbox{ and } g\,\). Show that the time after projection at which the direction of motion of \(P_1\) first makes an angle of \(45^{\circ}\) with the horizontal is \(\frac12 (1-\cot \alpha)T_1\,\). A particle \(P_2\) is projected under the same conditions. When the direction of the motion of \(P_2\) first makes an angle of \(45^{\circ}\) with the horizontal, the speed of \(P_2\) is instantaneously doubled. If \(T_2\) is the total flight time of \(P_2\), show that $$ \frac{2T_2}{T_1} = 1+\cot{\alpha} +\sqrt{1+3\cot^2{\alpha}} \;. $$

Show that the equation \[ ax^{2}+ay^{2}+2gx+2fy+c=0 \] where \(a>0\) and \(f^{2}+g^{2}>ac\) represents a circle in Cartesian coordinates and find its centre. The smooth and level parade ground of the First Ruritanian Infantry Division is ornamented by two tall vertical flagpoles of heights \(h_{1}\) and \(h_{2}\) a distance \(d\) apart. As part of an initiative test a soldier has to march in such a way that he keeps the angles of elevation of the tops of the two flagpoles equal to one another. Show that if the two flagpoles are of different heights he will march in a circle. What happens if the two flagpoles have the same height? To celebrate the King's birthday a third flagpole is added. Soldiers are then assigned to each of the three different pairs of flagpoles and are told to march in such a way that they always keep the tops of their two assigned flagpoles at equal angles of elevation to one another. Show that, if the three flagpoles have different heights \(h_{1},h_{2}\) and \(h_{3}\) and the circles in which the soldiers march have centres of \((x_{ij},y_{ij})\) (for the flagpoles of height \(h_{i}\) and \(h_{j}\)) relative to Cartesian coordinates fixed in the parade ground, then the \(x_{ij}\) satisfy \[ h_{3}^{2}\left(h_{1}^{2}-h_{2}^{2}\right)x_{12}+h_{1}^{2}\left(h_{2}^{2}-h_{3}^{2}\right)x_{23}+h_{2}^{2}\left(h_{3}^{2}-h_{1}^{2}\right)x_{31}=0, \] and the same equation connects the \(y_{ij}\). Deduce that the three centres lie in a straight line.

The Ruritanian army is supplied with shells which may explode at any time in flight but not before the shell reaches its maximum height. The effect of the explosion on any observer depends only on the distance between the exploding shell and the observer (and decreases with distance). Ruritanian guns fire the shells with fixed muzzle speed, and it is the policy of the gunners to fire the shell at an angle of elevation which minimises the possible damages to themselves (assuming the ground is level) - i.e. they aim so that the point on the descending trajectory that is nearest to them is as far away as possible. With that intention, they choose the angle of elevation that minimises the damage to themselves if the shell explodes at its maximum height. What angle do they choose? Does the shell then get any nearer to the gunners during its descent?