Problems

Two points \(A\) and \(B\) lie on horizontal ground. A particle \(P_1\) is projected from \(A\) towards \(B\) at an acute angle of elevation \(\alpha\) and simultaneously a particle \(P_2\) is projected from \(B\) towards \(A\) at an acute angle of elevation \(\beta\). Given that the two particles collide in the air a horizontal distance \(b\) from \(B\), and that the collision occurs after \(P_1\) has attained its maximum height \(h\), show that \[ 2h \cot\beta < b < 4h \cot\beta \hphantom{\,,} \] and \[ 2h \cot\alpha < a < 4h \cot\alpha \,, \] where \(a\) is the horizontal distance from \(A\) to the point of collision.

A smooth, straight, narrow tube of length \(L\) is fixed at an angle of \(30^\circ\) to the horizontal. A~particle is fired up the tube, from the lower end, with initial velocity \(u\). When the particle reaches the upper end of the tube, it continues its motion until it returns to the same level as the lower end of the tube, having travelled a horizontal distance \(D\) after leaving the tube. Show that \(D\) satisfies the equation \[ 4gD^2 - 2 \sqrt{3} \left( u^2 - Lg \right)D - 3L \left( u^2 - gL \right) = 0 \] and hence that \[ \frac{{\rm d}D}{ {\rm d}L} = - \frac{ 2\sqrt{3}gD - 3(u^2-2gL)} { 8gD - 2 \sqrt{3} \left(u^2 - gL \right)}. \] The final horizontal displacement of the particle from the lower end of the tube is \(R\). Show that \(\dfrac{\d R}{\d L} = 0\) when \(2D = L \sqrt 3\), and determine, in terms of \(u\) and \(g\), the corresponding value of \(R\).