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

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\frac{4}{3}

\frac{9}{5}

\frac{7}{3}

\frac{5}{2}

answer is A.

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Detailed Solution

Parabola equation is,

x^{2} = - 4 ay

Passing points is,

(6, - 4)

Put the values in equation, we get,

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

\Rightarrow 36 = 16 a \Rightarrow a = \frac{9}{4}

Thus, equation become,

x^{2} = - 9 y

Differentiating it with respect to x to find the slope,

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

(- 6, - 4), \frac{dy}{dx} ​ = - \frac{2}{9} \times (- 6) \frac{dy}{dx} = \frac{4}{3}

Thus, the slope can be given as,

tanθ = \frac{4}{3} ​; θ = ta n^{- 1} \frac{4}{3}

The angle of projection is

\frac{4}{3}

.
Option 1 is correct.

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