Question

A particle is projected train a point on the level ground and its height is $h$ when at horizontal distances $a$ and $2a$ from its point of projection. Find the velocity of projection.

Answer 1

If $v_0$ is the velocity of projection and $\alpha$ the angle of projection, the equation of trajectory is

$y=x\tan \alpha -\frac{1}{2}\frac{g{x}^2}{v_0^2{\cos }^2\alpha }$.............(i)

With origin at the point of projection,

$g{x}^2-2v_0^2\sin \alpha \cos \alpha .x+2v_0^2{\cos }^2\alpha .y=0$.........(ii)

Since the projectile passes through two points $(a,h)$ and $(2a,h)$, then $a$ and $2a$ must be roots of Eq (ii),

$a+2a=\frac{2v_0^2\sin \alpha \cos \alpha }{g}$.....................(iii)

and

$a\times 2a=\frac{2v_0^2{\cos }^2\alpha h}{g}$.................(iv)

Dividing Eqs (iii) by (iv), we get

$\frac{3a}{2a^2}=\frac{\tan \alpha }{h}$ or $\tan \alpha =\frac{3h}{2a}$

From Eq.(iv),

$v_0^2=\frac{g{a}^2}{h}se{c}^2\alpha =\frac{g{a}^2}{h}(1+ta{n}^2\alpha )=\frac{g{a}^2}{h}(1+\frac{9h^2}{4a^2})$

$=\frac{g}{4}(\frac{4a^2}{h}+9h)$ or $v_0=\frac{1}{2}\sqrt{(\frac{4a^2}{h}+9h)g}$