Question

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

Answer 1

The pictorial representation of the problem is shown below

Let the vertex be at origin. Since the parabola is open downwards its equation is $x^2=-4ay$.

It passes through $(6,-4)$.

$\Rightarrow 36=16a$

$\Rightarrow a=\frac{36}{16}$

$\Rightarrow a=\frac{9}{4}$

Thus the equation of the path becomes $x^2=-4\cdot \frac{9}{4}y$

$\Rightarrow x^2=-9y...\left(1\right)$

Differentiate $\left(1\right)$ with respect to $x$, we get

$2x=-9\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=-\frac{2x}{9}$

Hence slope $\frac{dy}{dx}=-\frac{2x}{9}$

Slope at $(-6,-4)$ is $\frac{dy}{dx}=-\frac{2(-6)}{9}=\frac{4}{3}$

That is, $tan\theta =\frac{4}{3}$

$\Rightarrow \theta =ta{n}^{-1}\left(\frac{4}{3}\right)$

Hence the angle of projection $\theta$ is $ta{n}^{-1}\left(\frac{4}{3}\right)$