4 problems found
In this question, the \(x\)- and \(y\)-axes are horizontal and the \(z\)-axis is vertically upwards.
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 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 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 \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]