Answer
Verified
439.8k+ views
Hint: In this particular question use the common fact that the rocket is always going upwards. And it is given it follows a parabolic path so the parabola should be vertically downwards so assume the equation of the parabola, ${x^2} = - 4ay$, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
The pictorial representation of the given problem is shown below
It is a common fact that the rocket is always going upwards.
And it is given it follows a parabolic path so the parabola should be vertically downwards.
So, let the equation of parabola
${x^2} = - 4ay..............\left( 1 \right)$
It is given that it reaches a maximum height 4m when it is 6m away from the starting point.
$\therefore $It is passing from$\left( { - 6, - 4} \right)$, so this point satisfies the equation of parabola.
From equation 1
$
{\left( { - 6} \right)^2} = - 4a\left( { - 4} \right) \Rightarrow 36 = 16a \Rightarrow a = \dfrac{9}{4} \\
\Rightarrow {x^2} = - 4\left( {\dfrac{9}{4}} \right)y = - 9y................\left( 2 \right) \\
$
Now we have to find out the angle of projection of the rocket.
So, we know angle of projection is the slope of the parabola at point $\left( { - 6, - 4} \right)$
Therefore slope${\text{ = }}\dfrac{{dy}}{{dx}} = \tan \theta $
From equation 2
\[y = \dfrac{{ - {x^2}}}{9}\]
Therefore differentiate the equation of parabola w.r.t.$x$
\[
\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{9}\left( {2x} \right) \\
\dfrac{{dy}}{{dx}}{\text{ at point }}\left( { - 6, - 4} \right) = \dfrac{{ - 1\left( {2 \times \left( { - 6} \right)} \right)}}{9} = \dfrac{{12}}{9} = \dfrac{4}{3} \\
\Rightarrow \tan \theta = \dfrac{4}{3} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right) \\
\]
So this is the required angle of projection.
Note: Whenever we face such type of problem the key point we have to remember is that the rocket is always goes up so the parabola is opening vertically downwards, then according to given conditions mark all the points and satisfy any point in the equation of parabola except (0, 0), so we get the value of $a$, then differentiate the equation of parabola and satisfying the starting points, then we will get the required angle of projection.
Complete step-by-step answer:
The pictorial representation of the given problem is shown below
It is a common fact that the rocket is always going upwards.
And it is given it follows a parabolic path so the parabola should be vertically downwards.
So, let the equation of parabola
${x^2} = - 4ay..............\left( 1 \right)$
It is given that it reaches a maximum height 4m when it is 6m away from the starting point.
$\therefore $It is passing from$\left( { - 6, - 4} \right)$, so this point satisfies the equation of parabola.
From equation 1
$
{\left( { - 6} \right)^2} = - 4a\left( { - 4} \right) \Rightarrow 36 = 16a \Rightarrow a = \dfrac{9}{4} \\
\Rightarrow {x^2} = - 4\left( {\dfrac{9}{4}} \right)y = - 9y................\left( 2 \right) \\
$
Now we have to find out the angle of projection of the rocket.
So, we know angle of projection is the slope of the parabola at point $\left( { - 6, - 4} \right)$
Therefore slope${\text{ = }}\dfrac{{dy}}{{dx}} = \tan \theta $
From equation 2
\[y = \dfrac{{ - {x^2}}}{9}\]
Therefore differentiate the equation of parabola w.r.t.$x$
\[
\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{9}\left( {2x} \right) \\
\dfrac{{dy}}{{dx}}{\text{ at point }}\left( { - 6, - 4} \right) = \dfrac{{ - 1\left( {2 \times \left( { - 6} \right)} \right)}}{9} = \dfrac{{12}}{9} = \dfrac{4}{3} \\
\Rightarrow \tan \theta = \dfrac{4}{3} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right) \\
\]
So this is the required angle of projection.
Note: Whenever we face such type of problem the key point we have to remember is that the rocket is always goes up so the parabola is opening vertically downwards, then according to given conditions mark all the points and satisfy any point in the equation of parabola except (0, 0), so we get the value of $a$, then differentiate the equation of parabola and satisfying the starting points, then we will get the required angle of projection.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE