5 problems found
A particle is projected from a point on a horizontal plane, at speed \(u\) and at an angle~\(\theta\) above the horizontal. Let \(H\) be the maximum height of the particle above the plane. Derive an expression for \(H\) in terms of \(u\), \(g\) and \(\theta\). A particle \(P\) is projected from a point \(O\) on a smooth horizontal plane, at speed \(u\) and at an angle~\(\theta\) above the horizontal. At the same instant, a second particle \(R\) is projected horizontally from \(O\) in such a way that \(R\) is vertically below \(P\) in the ensuing motion. A light inextensible string of length \(\frac12 H\) connects \(P\) and \(R\). Show that the time that elapses before the string becomes taut is \[ (\sqrt2 -1)\sqrt{H/g\,}\,. \] When the string becomes taut, \(R\) leaves the plane, the string remaining taut. Given that \(P\) and \(R\) have equal masses, determine the total horizontal distance, \(D\), travelled by \(R\) from the moment its motion begins to the moment it lands on the plane again, giving your answer in terms of \(u\), \(g\) and \(\theta\). Given that \(D=H\), find the value of \(\tan\theta\).
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.
Two particles \(P\) and \(Q\) are projected simultaneously from points \(O\) and \(D\), respectively, where~\(D\) is a distance \(d\) directly above \(O\). The initial speed of \(P\) is \(V\) and its angle of projection {\em above} the horizontal is \(\alpha\). The initial speed of \(Q\) is \(kV\), where \(k>1\), and its angle of projection {\em below} the horizontal is \(\beta\). The particles collide at time \(T\) after projection. Show that \(\cos\alpha = k\cos\beta\) and that \(T\) satisfies the equation \[ (k^2-1)V^2T^2 +2dVT\sin\alpha -d^2 =0\,. \] Given that the particles collide when \(P\) reaches its maximum height, find an expression for~\(\sin^2\alpha\) in terms of \(g\), \(d\), \(k\) and \(V\), and deduce that \[ gd\le (1+k)V^2\,. \]
A particle of unit mass is projected vertically upwards in a medium whose resistance is \(k\) times the square of the velocity of the particle. If the initial velocity is \(u\), prove that the velocity \(v\) after rising through a distance \(s\) satisfies \begin{equation*} v^{2}=u^{2}\e^{-2ks}+\frac{g}{k}(\e^{-2ks}-1). \tag{*} \end{equation*} Find an expression for the maximum height of the particle above he point of projection. Does equation \((*)\) still hold on the downward path? Justify your answer.
Solution: \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g - kv^2 \\ \Rightarrow && \int \frac{v}{g+kv^2} \d v &= \int -1 \d s \\ \Rightarrow && \frac{1}{2k}\ln(g+kv^2) &= -s + C \\ s =0, v = u: && \frac{1}{2k} \ln(g+ku^2) &= C \\ \Rightarrow && s &= \frac{1}{2k} \ln \frac{g+ku^2}{g+kv^2} \\ \Rightarrow && e^{-2ks} &= \frac{g+kv^2}{g+ku^2} \\ \Rightarrow && v^2 &= u^2e^{-2ks} + \frac{g}{k}(e^{-2ks}-1) \end{align*} The maximum height will be when \(v = 0\), ie \(\displaystyle s = \frac{1}{2k}\ln\left(1 + \frac{k}{g}u^2 \right)\). On the downward path the resistance will be going upwards, ie \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g + kv^2 \end{align*} but our solution is solving a different differential equation, therefore unless \(k=0\) the equation will be different.
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?