Question

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

Answer 1

Hint: In order to solve the problem first consider some general equation of the parabola and use the statement given in the problem as the point of parabola in order to find the constants. Finally in order to find the angle of projection use the equation of parabola.

Complete step-by-step answer:

The equation of parabola is of the form ${x^2} = - 4ay$

According to the given statement: rocket reaches a maximum height of 4mts when it is 6mts away from the point of projection.

This can be inferred as follows:

It passes through the point $\left( { - 6, - 4} \right)$

So the point must satisfy the equation of parabola

$

  \because {x^2} = - 4ay \\

   \Rightarrow {\left( { - 6} \right)^2} = - 4a\left( { - 4} \right) \\

   \Rightarrow 36 = 16a \\

   \Rightarrow a = \dfrac{{36}}{{16}} \\

   \Rightarrow a = \dfrac{9}{4} \\

 $

So we have got the value of a.

So the equation of parabola becomes

$

   \Rightarrow {x^2} = - 4\left( {\dfrac{9}{4}} \right)y \\

   \Rightarrow {x^2} = - 9y........(1) \\

 $

As we know that slope of the tangent for the curve is given by $\dfrac{{dy}}{{dx}}$
Now in order to find the angle of projection or $\tan \theta $ let us find $\dfrac{{dy}}{{dx}}$ for the curve or let us differentiate the curve with respect to x

\[

   \Rightarrow \dfrac{d}{{dx}}\left[ {{x^2}} \right] = \dfrac{d}{{dx}}\left[ { - 9y} \right]{\text{ or}} \\

   \Rightarrow \dfrac{d}{{dx}}\left[ { - 9y} \right] = \dfrac{d}{{dx}}\left[ {{x^2}} \right] \\
   \Rightarrow - 9\dfrac{d}{{dx}}\left[ y \right] = \dfrac{d}{{dx}}\left[ {{x^2}} \right] \\
   \Rightarrow - 9\dfrac{{dy}}{{dx}} = 2x{\text{ }}\left[ {\because \dfrac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}} \right] \\

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{9} \\

 \]

As we are having the point $\left( { - 6, - 4} \right)$ lying on the parabola. Let us substitute the value of x in order to find the numerical value of angle of projection.

$

  \because \dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{9} \\

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2\left( { - 6} \right)}}{9} = \dfrac{{12}}{9} \\

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{3} \\

 $

Now let us find the value of angle as $\tan \theta $ is same as $\dfrac{{dy}}{{dx}}$

$

   \Rightarrow \tan \theta = \dfrac{4}{3} \\

   \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{4}{3} \\

 $

Hence, the angle of projection for the given rocket is ${\tan ^{ - 1}}\dfrac{4}{3}$ .

Note: In order to solve such types of problems students must visualize the practical problem in terms of mathematical problems. The problem has become quite easier to solve after considering the path of the rocket as parabola. Students must remember that the slope of tangent to any curve is given by directly differentiating the curve equation with respect to x variable.