Problems

In this question, the \(x\)- and \(y\)-axes are horizontal and the \(z\)-axis is vertically upwards.

1. A particle \(P_\alpha\) is projected from the origin with speed \(u\) at an acute angle \(\alpha\) above the positive \(x\)-axis. The curve \(E\) is given by \(z = A - Bx^2\) and \(y = 0\). If \(E\) and the trajectory of \(P_\alpha\) touch exactly once, show that \[u^2 - 2gA = u^2(1 - 4AB)\cos^2\alpha\,.\] \(E\) and the trajectory of \(P_\alpha\) touch exactly once for all \(\alpha\) with \(0 < \alpha < \frac{1}{2}\pi\). Write down the values of \(A\) and \(B\) in terms of \(u\) and \(g\).

2. Describe the set of points which can be hit by particles from the explosion, explaining your answer.
3. Show that, at a time \(t\) after the explosion, the particles lie on a sphere whose centre and radius you should find.
4. Another particle \(Q\) is projected horizontally from the point \((0, 0, A)\) with speed \(u\) in the positive \(x\) direction. Show that, at all times, \(Q\) lies on the curve \(E\).
5. Show that for particles \(Q\) and \(P_\alpha\) to collide, \(Q\) must be projected a time \(\dfrac{u(1-\cos\alpha)}{g\sin\alpha}\) after the explosion.

Let \(L_a\) denote the line joining the points \((a,0)\) and \((0, 1-a)\), where \(0< a < 1\). The line \(L_b\) is defined similarly.

1. Determine the point of intersection of \(L_a\) and \(L_b\), where \(a\ne b\).
2. Show that this point of intersection, in the limit as \(b\to a\), lies on the curve \(C\) given by \[ y=(1-\sqrt x)^2\, \ \ \ \ (0< x < 1)\,. \]
3. Show that every tangent to \(C\) is of the form \(L_a\) for some \(a\).

A particle is projected from a point \(O\) on horizontal ground with initial speed \(u\) and at an angle of \(\theta\) above the ground. The motion takes place in the \(x\)-\(y\) plane, where the \(x\)-axis is horizontal, the \(y\)-axis is vertical and the origin is \(O\). Obtain the Cartesian equation of the particle's trajectory in terms of \(u\), \(g\) and~\(\lambda\), where \(\lambda=\tan\theta\). Now consider the trajectories for different values of \(\theta\) with \(u\)~fixed. Show that for a given value of~\(x\), the coordinate~\(y\) can take all values up to a maximum value,~\(Y\), which you should determine as a function of \(x\), \(u\) and~\(g\). Sketch a graph of \(Y\) against \(x\) and indicate on your graph the set of points that can be reached by a particle projected from \(O\) with speed \(u\). Hence find the furthest distance from \(O\) that can be achieved by such a projectile.

1. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (y')^2 -xy'+y=0\,. \tag{\(*\)} \] By differentiating \((*)\) with respect to \(x\), show that either \(y''=0\) or \(2y'=x\,\). Hence show that the curve is either a straight line of the form \(y=mx+c\), where \(c=-m^2\), or the parabola \(4y=x^2\).
2. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (x^2-1)(y')^2 -2xyy'+y^2-1=0\,. \] Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.

A particle is projected from a point \(O\) with speed \(\sqrt{2gh},\) where \(g\) is the acceleration due to gravity. Show that it is impossible, whatever the angle of projection, for the particle to reach a point above the parabola \[ x^{2}=4h(h-y), \] where \(x\) is the horizontal distance from \(O\) and \(y\) is the vertical distance above \(O\). State briefly the simplifying assumptions which this solution requires.