Question

A particle is projected from point G from the ground such that it touches the points B,C,D and E of a regular hexagon of side root12 meters as shown in figure The maximum height reached by the projectile from the ground(in meters) is

Answer 1

Let I be the midpoint of AF and J be the vertical projection of the point B along the horizontal.

Here, $\sf{AF=\sqrt{12}\ m,}$ so $\sf{AI=IF=\dfrac{\sqrt{12}}{2}\ m.}$

We see $\sf{GI=IH=\dfrac{R}{2}.}$ So,

$\longrightarrow\sf{AG=GI-AI}$

$\longrightarrow\sf{AG=\dfrac{R-\sqrt{12}}{2}}$

From triangle AFC,

$\longrightarrow\sf{AC=AF\tan60^o}$

$\longrightarrow\sf{AC=\sqrt{12}\times\sqrt3}$

$\longrightarrow\sf{AC=\sqrt{36}\ m}$

$\longrightarrow\sf{AC=6\ m}$

We know the equation of the trajectory of a projectile is,

$\longrightarrow\sf{y=x\tan\theta\left[1-\dfrac{x}{R}\right]}$

When the projectile reaches the point C, its horizontal displacement is AG and vertical displacement is AC. So, by equation of trajectory,

$\longrightarrow\sf{AC=AG\tan\theta\left[1-\dfrac{AG}{R}\right]}$

$\longrightarrow\sf{6=\dfrac{R-\sqrt{12}}{2}\left[1-\dfrac{R-\sqrt{12}}{2R}\right]\tan\theta}$

$\longrightarrow\sf{6=\dfrac{R-\sqrt{12}}{2}\left[\dfrac{2R-R+\sqrt{12}}{2R}\right]\tan\theta}$

$\longrightarrow\sf{6=\dfrac{(R-\sqrt{12})(R+\sqrt{12})}{4R}\tan\theta}$

$\longrightarrow\sf{6=\dfrac{R^2-12}{4R}\,\tan\theta}$

$\longrightarrow\sf{\dfrac{\tan\theta}{4}=\dfrac{6R}{R^2-12}\quad\quad\dots(1)}$

But, for a projectile it is true that,

$\longrightarrow\sf{\dfrac{H}{R}=\dfrac{\tan\theta}{4}\quad\quad\dots(2)}$

Putting (2) in (1),

$\longrightarrow\sf{\dfrac{H}{R}=\dfrac{6R}{R^2-12}}$

$\longrightarrow\sf{H=\dfrac{6R^2}{R^2-12}\quad\quad\dots(3)}$

From triangle ABJ,

$\longrightarrow\sf{BJ=AB\sin60^o}$

$\longrightarrow\sf{BJ=\sqrt{12}\times\dfrac{\sqrt3}{2}}$

$\longrightarrow\sf{BJ=3\ m}$

And,

$\longrightarrow\sf{AJ=AB\cos60^o}$

$\longrightarrow\sf{AJ=\dfrac{\sqrt{12}}{2}\ m}$

Thus,

$\longrightarrow\sf{GJ=GI-(AI+AJ)}$

$\longrightarrow\sf{GJ=\dfrac{R}{2}-\left(\dfrac{\sqrt{12}}{2}+\dfrac{\sqrt{12}}{2}\right)}$

$\longrightarrow\sf{GJ=\dfrac{R-2\sqrt{12}}{2}}$

When the projectile reaches the point B, its horizontal displacement is GJ and vertical displacement is JB. So, by equation of trajectory,

$\longrightarrow\sf{BJ=GJ\tan\theta\left[1-\dfrac{GJ}{R}\right]}$

$\longrightarrow\sf{3=\dfrac{R-2\sqrt{12}}{2}\left[1-\dfrac{R-2\sqrt{12}}{2R}\right]\tan\theta}$

$\longrightarrow\sf{3=\dfrac{R-2\sqrt{12}}{2}\left[\dfrac{2R-R+2\sqrt{12}}{2R}\right]\tan\theta}$

$\longrightarrow\sf{3=\dfrac{\left(R-2\sqrt{12}\right)\left(R+2\sqrt{12}\right)}{4R}\tan\theta}$

$\longrightarrow\sf{3=\dfrac{R^2-48}{4R}\,\tan\theta}$

$\longrightarrow\sf{\dfrac{\tan\theta}{4}=\dfrac{3R}{R^2-48}}$

From (2),

$\longrightarrow\sf{\dfrac{H}{R}=\dfrac{3R}{R^2-48}}$

$\longrightarrow\sf{H=\dfrac{3R^2}{R^2-48}\quad\quad\dots(4)}$

Equating (3) and (4),

$\longrightarrow\sf{\dfrac{6R^2}{R^2-12}=\dfrac{3R^2}{R^2-48}}$

$\longrightarrow\sf{\dfrac{2}{R^2-12}=\dfrac{1}{R^2-48}}$

$\longrightarrow\sf{2R^2-96=R^2-12}$

$\longrightarrow\sf{R^2=84}$

Then from (4),

$\longrightarrow\sf{H=\dfrac{3\times84}{84-48}}$

$\longrightarrow\underline{\underline{\sf{H=7\ m}}}$

Hence 7 m is the answer.