51 problems found
A particle is projected over level ground with a speed \(u\) at an angle \(\theta\) above the horizontal. Derive an expression for the greatest height of the particle in terms of \(u\), \(\theta\) and \(g\). A particle is projected from the floor of a horizontal tunnel of height \({9\over 10}d\). Point \(P\) is \({1\over 2}d\) metres vertically and \(d\) metres horizontally along the tunnel from the point of projection. The particle passes through point \(P\) and lands inside the tunnel without hitting the roof. Show that \[ \arctan \textstyle {3 \over 5} < \theta < \arctan \, 3 \;. \]
Solution: \begin{align*} && v^2 &= u^2 + 2as \\ (\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\ \Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g} \end{align*} To avoid hitting the ceiling \begin{align*} && \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\ \Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\ \end{align*} In order to pass through \(P\) we need \begin{align*} && d &= u \cos \theta t \\ && \frac12 d &= u \sin \theta t - \frac12 g t^2 \\ \Rightarrow && \frac12 &= \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\ &&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\ \Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\ &&&= (5\tan \theta - 3)(\tan \theta - 3) \\ \\ \Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\ && \theta &= \left (\arctan \tfrac35, \arctan 3 \right) \end{align*}
A particle is projected with speed \(V\) at an angle \(\theta\) above the horizontal. The particle passes through the point \(P\) which is a horizontal distance \(d\) and a vertical distance \(h\) from the point of projection. Show that \[ T^2 -2kT + \frac{2kh}{d}+1=0\;, \] where \(T=\tan\theta\) and \(\ds k= \frac{V^2}{gd}\,\). %Derive an equation relating \(\tan \theta\), \(V\), \(g\), \(d\) and \(h\). Show that, if \(\displaystyle {kd > h + \sqrt {h^2 + d^2}}\;\), there are two distinct possible angles of projection. Let these two angles be \(\alpha\) and \(\beta\). Show that \(\displaystyle \alpha + \beta = \pi - \arctan ( {d/ h}) \,\).
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 $$ \frac{2T_2}{T_1} = 1+\cot{\alpha} +\sqrt{1+3\cot^2{\alpha}} \;. $$
A particle is projected from a point \(O\) on a horizontal plane
with speed \(V\) and at an angle
of elevation \(\alpha\). The vertical plane in which the motion takes place
is perpendicular to two vertical walls, both of height \(h\), at distances
\(a\) and \(b\) from \(O\). Given that the particle just passes over the
walls, find \(\tan\alpha\) in terms of \(a\), \(b\) and \(h\) and
show that
\[
\frac{2V^2} g = \frac {ab} h +\frac{ (a+b)^2 h}{ab} \;.
\]
The heights of the walls are now increased by the same small positive
amount \(\delta h\,\).
A second particle is projected so that it just passes over
both walls, and the new angle and speed of projection
are \(\alpha +\delta \alpha \) and \(V+\delta V\), respectively.
Show that
\[
\sec^2 \alpha \, \delta \alpha \approx \frac {a+b}{ab}\,\delta h \;,
\]
and deduce that \(\delta \alpha >0\,\). Show also that
\(\delta V\) is positive if \(h> ab/(a+b)\) and negative if \(h
A gun is sited on a horizontal plain and can fire shells in any direction and at any elevation at speed \(v\). The gun is a distance \(d\) from a straight railway line which crosses the plain, where \(v^2>gd\). The gunner aims to hit the line, choosing the direction and elevation so as to maximize the time of flight of the shell. Show that the time of flight, \(T\), of the shell satisfies \[ %\frac{2v}{g} \sin \left( \frac12 \arccos \frac{gd}{v^2}\right)\,. g^2 T^2 = 2 v^2 +2 \left(v^4 -g^2d^2\right)^{\frac12}\,. \] Extension: (Not in original paper) Find the time of flight if the gun is constrained so that the angle of elevation \(\alpha \) is not greater than \( 45^\circ\).
Solution: If we fire the gun at an angle to the track, as long as we can travel a horizontal distance \(\geq d\) we can hit the track. Suppose we am at an elevatation \(\alpha\), then \begin{align*} (\uparrow): && s &= ut + \frac12 at^2 \\ && 0 &= v\sin \alpha T - \frac12 g T^2 \\ \Rightarrow &&T &= \frac{2v\sin \alpha}{g}\\ \\ (\rightarrow): && s &= ut \\ && s &= v \cos \alpha T \\ &&&= v\sqrt{1-\sin^2 \alpha} T \\ &&&= vT\sqrt{1 - \frac{g^2T^2}{4v^2}}\\ &&&= \frac{T}{2}\sqrt{4v^2-g^2T^2}\\ \Rightarrow && d & \leq \frac{T}{2}\sqrt{4v^2-g^2T^2} \\ \Rightarrow && 4g^2d^2&\leq g^2T^2(4v^2-g^2T^2) \\ \Rightarrow && 0 &\leq -(g^2T^2)^2 + 4v^2 (g^2T^2)-4g^2d^2 \\ &&&=4v^4-4g^2d^2 -\left (g^2T^2-2v^2 \right)^2 \\ \Rightarrow && \left (g^2T^2-2v^2 \right)^2 & \leq 4v^4-4g^2d^2 \\ \Rightarrow && g^2T^2 &\leq 2v^2+2\sqrt{v^4-g^2d^2} \end{align*} Therefore the maximum value for \(g^2T^2\) is \(2v^2+2\sqrt{v^4-g^2d^2}\) Notice that we are hitting the track directly at \(d\). This is because to maximise the time of flight (for a fixed speed) we want to maximise the angle of elevation. Therefore we want the highest angle where we still hit the track (which is clearly the shortest distance). If we are constraint to \(\alpha \leq 45^\circ\) we know that \(T\) is maximised when \(\alpha = 45^\circ\) (and we will reach the track since the range \(\frac{v^2 \sin 2 \alpha}{g}\) is increasing). Therefore the maximum time is \(T = \frac{\sqrt{2}v}{g}\)
A two-stage missile is projected from a point \(A\) on the ground with horizontal and vertical velocity components \(u\) and \(v\), respectively. When it reaches the highest point of its trajectory an internal explosion causes it to break up into two fragments. Immediately after this explosion one of these fragments, \(P\), begins to move vertically upwards with speed \(v_e\), but retains the previous horizontal velocity. Show that \(P\) will hit the ground at a distance \(R\) from \(A\) given by $$ \frac{gR}u = v+v_e + \sqrt{v_e^2 +v^2}\, . $$ It is required that the range \(R\) should be greater than a certain distance \(D\) (where \(D> 2uv/g\)). Show that this requirement is satisfied if \[ v_e> \frac{gD}{2u}\left( \frac{gD-2uv}{gD-uv}\right). \] \noindent[{\sl The effect of air resistance is to be neglected.}]
A child is playing with a toy cannon on the floor of a long railway carriage. The carriage is moving horizontally in a northerly direction with acceleration \(a\). The child points the cannon southward at an angle \(\theta\) to the horizontal and fires a toy shell which leaves the cannon at speed \(V\). Find, in terms of \(a\) and \(g\), the value of \(\tan 2\theta\) for which the cannon has maximum range (in the carriage). If \(a\) is small compared with \(g\), show that the value of \(\theta\) which gives the maximum range is approximately \[ \frac \pi 4 + \frac a {2g}, \] and show that the maximum range is approximately \(\displaystyle \frac {V^2} g + \frac {V^2a}{g^2}. \)
Solution: \begin{align*} && s_x &= V \cos \theta t + \frac12at^2 \\ && s_y &= V \sin \theta t - \frac12 gt^2 \\ \Rightarrow && T &= \frac{2V \sin \theta}g \\ \Rightarrow && s_{max} &= \frac{2V^2 \sin \theta \cos \theta}{g} + \frac12a \frac{4V^2 \sin^2 \theta}{g^2} \\ &&&= (g \sin 2 \theta+2a\sin^2 \theta)\frac{V^2}{g^2} \\ && \frac{\d s_{max}}{\d \theta} &= (2g \cos 2 \theta +4 a \cos \theta \sin \theta)\frac{V^2}{g^2} \\ &&&= (2g \cos 2\theta + 2a \sin2 \theta) \frac{V^2}{g^2} \\ \Rightarrow && \tan 2\theta &= -\frac{a}{g} \\ \Rightarrow && 2 \theta &\in (\frac{\pi}2, \pi) \\ \Rightarrow && \tan \left (\frac{\pi}{2} - 2 \theta\right) &=-\frac{a}{g} \\ \Rightarrow && \frac{\pi}{2} - 2 \theta&\approx -\frac{a}{g} \\ \Rightarrow && \theta &\approx \frac{\pi}{4} + \frac{a}{2g} \\ \\ && s_{max} & \approx \left (g \sin \left (\frac{\pi}{2} + \frac{a}{g} \right)+2a\sin^2 \left ( \frac{\pi}{4} + \frac{a}{2g}\right)\right)\frac{V^2}{g^2} \\ &&&\approx \left (g \cdot 1+2a\left( \frac{1}{\sqrt{2}}(\frac{a}{2g}+1)\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\left(1+\frac{a}{g}\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\right)\frac{V^2}{g^2} \\ &&&= \frac{V^2}{g} + \frac{V^2a}{g} \end{align*}
A shell explodes on the surface of horizontal ground. Earth is scattered in all directions with varying velocities. Show that particles of earth with initial speed \(v\) landing a distance \(r\) from the centre of explosion will do so at times \(t\) given by \[ {\textstyle \frac{1}{2}} g^2t^2=v^{2}\pm\surd(v^{4}-g^{2}r^{2}). \] Find an expression in terms of \(v\), \(r\) and \(g\) for the greatest height reached by such particles.
A fielder, who is perfectly placed to catch a ball struck by the batsman in a game of cricket, watches the ball in flight. Assuming that the ball is struck at the fielder's eye level and is caught just in front of her eye, show that \(\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) \) is constant, where \(\theta\) is the angle between the horizontal and the fielder's line of sight. In order to catch the next ball, which is also struck towards her but at a different velocity, the fielder runs at constant speed \(v\) towards the batsman. Assuming that the ground is horizontal, show that the fielder should choose \(v\) so that \(\frac{ {\rm d}}{{\rm d t}} (\tan\theta ) \) remains constant.
Solution: Set up a coordinate frame such that the position of the catch is the origin and the time of the catch is \(t = 0\) . We must have then that the trajectory of the ball is \(\mathbf{s} =\mathbf{u} t + \frac12 \mathbf{g} t^2 = \binom{u_x t}{u_y t - \frac12 gt^2}\). We must then have: \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{u_x t} \\ &&&= \frac{u_y}{u_x} - \frac{g}{2u_x} t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= 0 - \frac{g}{2 u_x} \end{align*} which is clearly constant. Set coordinates so \(y\)-axis starts from eye-level and \(t = 0\) the first time the ball reaches that level. (Or move the trajectory backwards if that's not the case). Then the ball has trajectory \(\binom{u_xt}{u_yt - \frac12 gt^2}\). The ball reaches eye level a second time when \(t = \frac{2u_y}{g}\), ie at a point \(\frac{2u_xu_y}{g}\). The fielder therefore needs to have position \(f + (u_x-\frac{g}{2u_y}f)t\) at all times. Therefore \begin{align*} && \tan \theta &= \frac{u_y t - \frac12 gt^2}{f + (u_x-\frac{g}{2u_y}f)t - u_x t} \\ &&&= \frac{u_y t - \frac12 gt^2}{f(1-\frac{g}{2u_y}t)} \\ &&&= \frac{u_yt ( 1- \frac{g}{2u_y}t)}{f( 1- \frac{g}{2u_y}t)} \\ &&&= u_y t \\ \Rightarrow && \frac{\d}{\d \theta} \left ( \tan \theta \right) &= u_y \end{align*} Ie \( \frac{\d}{\d \theta} \left ( \tan \theta \right) \) is constant as required.
A tennis player serves from height \(H\) above horizontal ground, hitting the ball downwards with speed \(v\) at an angle \(\alpha\) below the horizontal. The ball just clears the net of height \(h\) at horizontal distance \(a\) from the server and hits the ground a further horizontal distance \(b\) beyond the net. Show that $$v^2 = \frac{ g(a+b)^2(1+\tan^2\alpha)}{ 2[H-(a+b)\tan\alpha]}$$ and $$\tan\alpha = \frac{2a+b }{ a(a+b)}H - \frac{a+b }{ ab}h \,.$$ By considering the signs of \(v^2\) and \(\tan\alpha\), find upper and lower bounds on \(H\) for such a serve to be possible.
Solution: We have \begin{align*} \rightarrow: && a &= v\cos \alpha t_{net} \\ \Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\ \downarrow: && H-h &= v\sin \alpha t_{net} + \frac12 g t_{net}^2 \\ &&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\ &&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\ \\ \rightarrow: && a+b &= v \cos \alpha t_{ground} \\ && t_{ground} &= \frac{a+b}{v \cos \alpha}\\ \downarrow: && H &= v\sin \alpha t_{ground} + \frac12 g t_{ground}^2 \\ &&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\ \\ (**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\ (a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\ \Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\ \Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h \end{align*} Noting that \(v^2 \geq 0\) and the numerator is positive, we must have \begin{align*} && H &> (a+b)\tan \alpha \\ &&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\ \Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\ \Rightarrow && H &< \frac{a+b}{b} h \end{align*} Noting that \(\tan \alpha > 0\) we must have \begin{align*} && \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\ \Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h \end{align*}
A particle is projected under the influence of gravity from a point \(O\) on a level plane in such a way that, when its horizontal distance from \(O\) is \(c\), its height is \(h\). It then lands on the plane at a distance \(c+d\) from \(O\). Show that the angle of projection \(\alpha\) satisfies \[ \tan\alpha=\frac{h(c+d)}{cd} \] and that the speed of projection \(v\) satisfies \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]
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.
Solution: The position of the particle is projected at angle \(\theta\) is \((x,y) = (v \cos \theta t, v \sin \theta t - \frac12 g t^2)\), ie \(t = \frac{x}{v \cos \theta}\), \begin{align*} && y &= x\tan \theta -\frac12 g \frac{x^2}{v^2} \sec^2 \theta \\ && y &= x \tan \theta -\frac{1}{4h} (1+\tan^2 \theta) x^2 \\ && 0 &= \frac{1}{4h} x^2\tan^2 \theta - x \tan \theta + \frac{x^2}{4h} +y \\ \Delta \geq 0: && 0 &\leq \Delta = x^2-4\frac{x^2}{4h}\left (\frac{x^2}{4h}+y \right) \\ &&&=1-\frac{1}{4h^2}(x^2+4hy) \\ \Rightarrow && x^2+4hy &\leq 4h^2 \\ \Rightarrow && x^2 &\leq 4h(h-y) \end{align*} We are assuming that there are no forces acting other than gravity (eg air resistance)
A cannon is situated at the bottom of a plane inclined at angle \(\beta\) to the horizontal. A (small) cannon ball is fired from the cannon at an initial speed \(u.\) Ignoring air resistance, find the angle of firing which will maximise the distance up the plane travelled by the cannon ball and show that in this case the ball will land at a distance \[ \frac{u^{2}}{g(1+\sin\beta)} \] from the cannon.
A cannon-ball is fired from a cannon at an initial speed \(u\). After time \(t\) it has reached height \(h\) and is at a distance \(\sqrt{x^{2}+h^{2}}\) from the cannon. Ignoring air resistance, show that \[ \tfrac{1}{4}g^{2}t^{4}-(u^{2}-gh)t^{2}+h^{2}+x^{2}=0. \] Hence show that if \(u^{2}>2gh\) then the horizontal range for a given height \(h\) and initial speed \(u\) is less than or equal to \[ \frac{u\sqrt{u^{2}-2gh}}{g}. \] Show that there is always an angle of firing for which this value is attained.
Solution: Suppose it is fired with angle to the horizontal \(\alpha\), then \begin{align*} \rightarrow: && x &= u\cos \alpha \cdot t \\ \uparrow: && h &= u \sin \alpha \cdot t - \frac12 g t^2 \\ \Rightarrow && u\cos \alpha &= \frac{x}{t} \\ && u \sin \alpha &= \frac{h + \frac12 gt^2}{t} \\ \Rightarrow && u^2 &= \frac{x^2}{t^2} + \frac{(h + \frac12 gt^2)^2}{t^2} \\ \Rightarrow && 0 &= x^2+h^2-u^2t^2+ght^2+\tfrac14 g^2 t^4 \\ &&&= \tfrac14 g^2 t^4 - (u^2 - gh)t^2 + h^2 + x^2 \end{align*} For a distance \(x\) to be achievable there must be a root to this quadratic in \(t^2\), ie \begin{align*} && 0 &\leq \Delta = (u^2-gh)^2 - 4 \cdot \tfrac14 g^2 (h^2 + x^2) \\ \Rightarrow && x^2 &\leq \frac{(u^2-gh)^2}{g^2} - h^2 \\ &&&= \frac{u^4+g^2h^2 - 2ghu^2-g^2h^2}{g^2} \\ &&&= \frac{u^2(u^2-2gh)}{g^2} \\ \Rightarrow && x &\leq \frac{u\sqrt{u^2-2gh}}{g} \end{align*} This is achieved when \begin{align*} && t^2 &= \frac{u^2-gh}{\tfrac12g^2}\\ &&&= \frac{2(u^2-gh)}{g^2} \\ \Rightarrow && \cos \alpha &= \frac{u\sqrt{u^2-2gh}}{g} \cdot \frac{g}{\sqrt{2(u^2-gh)}} \frac{1}{u} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} ie when \(\alpha = \frac{\pi}{4}\)
As part of a firework display a shell is fired vertically upwards with velocity \(v\) from a point on a level stretch of ground. When it reaches the top of its trajectory an explosion it splits into two equal fragments each travelling at speed \(u\) but (since momentum is conserved) in exactly opposite (not necessarily horizontal) directions. Show, neglecting air resistance, that the greatest possible distance between the points where the two fragments hit the ground is \(2uv/g\) if \(u\leqslant v\) and \((u^{2}+v^{2})/g\) if \(v\leqslant u.\)
Solution: Since \(v^2 - u^2 = 2as\) we have the initial height reached is \(\frac{v^2}{2g}\). At the point of explosion, the velocities are \(\pm \binom{u \cos \theta}{u \sin \theta}\) where \(0 \leq \theta < \frac{\pi}{2}\). Looking vertically: \begin{align*} && -\frac{v^2}{2g} &= \pm u \sin \theta t - \frac12gt^2 \\ \Rightarrow && t &= \frac{\mp u \sin \theta \pm \sqrt{u^2 \sin^2 \theta - 4 \cdot \left (-\frac12 g \right) \cdot (\frac{v^2}{2g})}}{2(-\frac12g)} \\ &&&= \frac{\pm u \sin \theta \mp \sqrt{u^2 \sin^2 \theta+v^2}}{g}\\ &&&= \frac{\pm u \sin \theta +\sqrt{u^2 \sin^2 \theta+v^2}}{g} \end{align*} Since we always want the positive \(t\). Then the horizontal distance travelled will be \begin{align*} && s &= u \cos \theta (t_1 + t_2) \\ &&&= u \cos \theta \frac{2\sqrt{u^2 \sin^2 \theta+v^2}}{g} \\ &&&= \frac{2u \cos \theta \sqrt{u^2 \sin^2 \theta + v^2}}{g} \\ &&s^2 &= \frac{4u^2}{g^2} \cos^2 \theta ({u^2 \sin^2 \theta + v^2}) \\ &&&= \frac{4u^2}{g^2} \left (-u^2\cos^4 \theta + (v^2+u^2)\cos^2 \theta \right) \\ &&&= \frac{4u^2}{g^2} \left (-u^2\left ( \cos^2 \theta - \frac{v^2+u^2}{2u^2}\right)^2 + \frac{(v^2+u^2)^2}{4u^2} \right) \\ &&&= \frac{(v^2+u^2)^2}{g^2} - \frac{4u^4}{g^2}\left ( \cos^2 \theta - \frac{v^2+u^2}{2u^2}\right)^2 \end{align*} If \(u \geq v\) then such a \(\theta\) exists such that we can achieve the maximum, ie \(s = \frac{v^2+u^2}{g}\). If not, then we will achieve our maximum when \(\cos \theta = 1\), ie \(\sin \theta = 0\) and \(s = \frac{2uv}{g}\)