Question

On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of $4m$ when it is $6m$ away from the point of projection. Finally it reaches the ground $12m$ away from the salting point. Find the angle of projection.

Answer 1

The equation of the parabola is of the form $x^2=-4ay$
It passes through $(6,-4)$
$\therefore$ $36=16a\Rightarrow$ $a=\frac{9}{4}$
The equation is $x^2=-9y$ .....$\left(1\right)$
To find the solpe at $(-6,-4)$
Differentiating $\left(1\right)$ w.r.t $x$, we get
$2x=-9\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{-2}{9}x$
at $(-6,-4)$, $\frac{dy}{dx}=\frac{-2}{9}\times (-6)=\frac{4}{3}$
(i.e.,) $\tan \theta =\frac{4}{3};\theta ={\tan }^{-1}\frac{4}{3}$
$\therefore$ the angle of projection is ${\tan }^{-1}\frac{4}{3}$.