Compare Problem Quality

A tennis ball is projected from a height of \(2h\) above horizontal ground with speed \(u\) and at an angle of \(\alpha\) below the horizontal. It travels in a plane perpendicular to a vertical net of height \(h\) which is a horizontal distance of \(a\) from the point of projection. Given that the ball passes over the net, show that \[ \frac 1{u^2}< \frac {2(h-a\tan\alpha)}{ga^2\sec^2\alpha}\,. \] The ball lands before it has travelled a horizontal distance of \(b\) from the point of projection. Show that \[ \sqrt{u^2\sin^2\alpha +4gh \ } < \frac{bg}{u\cos\alpha} + u \sin\alpha\,. \] Hence show that \[ \tan\alpha < \frac{h(b^2-2a^2)}{ab(b-a)}\,. \]

---

Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

The numbers \(x,y\) and \(z\) are non-zero, and satisfy \[ 2a-3y=\frac{\left(z-x\right)^{2}}{y}\quad\mbox{ and }\quad2a-3z=\frac{\left(x-y\right)^{2}}{z}, \] for some number \(a\). If \(y\neq z\), prove that \[ x+y+z=a, \] and that \[ 2a-3x=\frac{\left(y-z\right)^{2}}{x}. \] Determine whether this last equation holds only if \(y\neq z\).

---

Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).