3 problems found
{\sl In this question take the acceleration due to gravity to be \(10\,{\rm m \,s}^{-2}\) and neglect air resistance.} The point \(O\) lies in a horizontal field. The point \(B\) lies \(50\,\)m east of \(O\). A particle is projected from \(B\) at speed \(25\,{\rm m\,s}^{-1}\) at an angle \(\arctan \frac12\) above the horizontal and in a direction that makes an angle \(60^\circ\) with \(OB\); it passes to the north of \(O\).
Point \(B\) is a distance \(d\) due south of point \(A\) on a horizontal plane. Particle \(P\) is at rest at \(B\) at \(t=0\), when it begins to move with constant acceleration \(a\) in a straight line with fixed bearing~\(\beta\,\). Particle \(Q\) is projected from point \(A\) at \(t=0\) and moves in a straight line with constant speed \(v\,\). Show that if the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\), then \[ v^2 \ge ad \l 1 - \cos \beta \r\;. \] Show further that if \(v^2 >ad(1-\cos\beta)\) then the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\) before \(P\) has moved a distance \(d\,\).
The points \(A,B\) and \(C\) lie on the surface of the ground, which is an inclined plane. The point \(B\) is 100m due north of \(A,\) and \(C\) is 60m due east of \(B\). The vertical displacements from \(A\) to \(B,\) and from \(B\) to \(C\), are each 5m downwards. A plane coal seam lies below the surface and is to be located by making vertical bore-holes at \(A,B\) and \(C\). The bore-holes strike the coal seam at 95m, 45m and 76m below \(A,B\) and \(C\) respectively. Show that the coal seam is inclined at \(\cos^{-1}(\frac{4}{5})\) to the horizontal. The coal seam comes to the surface along a line. Find the bearing of this line.
Solution: Set up a coordinate system so that \(x\) is E-W, \(y\) is N-S and \(z\) is the vertical direction. Also assume \(B\) is the origin, then, \(A = \begin{pmatrix} 0 \\ -100 \\ 5\end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}, C= \begin{pmatrix} 60 \\ 0\\ -5\end{pmatrix},\). The coal seam has points: \(\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}, \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix},\) Therefore we can find the normal to the coal seam: \begin{align*} \mathbf{n} &= \left (\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \times \left ( \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \\ &= \begin{pmatrix} 0 \\ - 100 \\ -45\end{pmatrix} \times \begin{pmatrix} 60 \\ 0 \\ -36\end{pmatrix} \\ &= \begin{pmatrix} 3600 \\ -60 \cdot 45 \\ 60 \cdot 100 \end{pmatrix} \\ &= 300\begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \end{align*} To measure the incline \(\theta\) to the horizontal we can take a dot with \(\hat{\mathbf{k}}\), to see: \begin{align*} \cos \theta &= \frac{20}{\sqrt{12^2+(-9)^2+20^2} \sqrt{1^2+0^2+0^2}} \\ &= \frac{20}{25} \\ &= \frac{4}{5} \end{align*} Therefore the angle is \(\cos^{-1} \tfrac 45\) The equation of the seam is \(12x - 9y + 20z = -900\). The equation of the surface is \(5x + 3y + 60z = 0\) We can compute the direction of the overlap again with a cross product: \begin{align*} \mathbf{d} &= \begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \times \begin{pmatrix} 5 \\ 3 \\ 60\end{pmatrix} \\ &= \begin{pmatrix} -600 \\ -620 \\ 81 \end{pmatrix} \end{align*} To get the bearing of this vector we just need to look at the \(x\) and \(y\) components, so it will be \(\tan^{-1} \frac{600}{620} = \tan^{-1} \frac{30}{31}\)