Question

A particle projected at a definite angle $\alpha$ to the horizontal passes through points $(a,b)$ and $(b,a)$, referred to horizontal and vertical axes through the point of projection. Show that:
a. The horizontal range $R=\frac{a^2+ab+b^2}{a+b}$
b. The angle of projection $\alpha$ is given by
$ta{n}^{-1}\left[\frac{a^2+ab+b^2}{ab}\right]$

Answer 1

a. The equation of the trajectory of the particle

$y=x\tan \alpha (1-\frac{x}{R})$ ...(i)

Since the particle passes through the points with coordinates $(a,b)$ and $(b,a)$, they should satisfy the equation of the curve.

$b=a\tan \alpha (1-\frac{1}{R})$ and (ii) $a=b\tan \alpha (1-\frac{b}{R})$ ...(ii)

Dividing $\frac{b^2}{a^2}=\frac{(1-\frac{a}{R})}{(1-\frac{b}{R})}$

$\Rightarrow b^2-\frac{b^3}{R}=a^2-\frac{a^3}{R}$

$\Rightarrow \frac{1}{R}[a^3-b^3]=a^2-b^2$

$\Rightarrow R=\frac{a^3-b^3}{a^2-b^2}=\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}=\frac{a^2+ab+b^2}{a+b}$

Hence, proved. [$\therefore a\ne b$]

b. Substituting the expression for $R$ in (ii),

$=\frac{b}{a}=tan\alpha [1-\frac{a(a+b)}{a^2+ab+b^2}]$

$=tan\alpha \left[\frac{a^2+ab+b^2-a^2-ab}{a^2+ab+b^2}\right]$

$tan\alpha =\frac{a^2+ab+b^2}{ab}\Rightarrow \alpha =ta{n}^{-1}\left[\frac{a^2+ab+b^2}{ab}\right]$

Hence proved.