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Exercise 5.5 · Q9

Q.On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4 m4\,\text m when it is 6 m6\,\text m away from the point of projection. Finally it reaches the ground 12 m12\,\text m away from the starting point. Find the angle of projection.

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The maximum height point is the vertex; the start and landing points confirm it (both give a downward parabola symmetric about x=6x=6); then differentiate to get the launch slope, i.e. the tangent of the projection angle.

Step 1. Set up. Vertex (6,4)(6,4) (max height, 6 m6\,\text m out), opens down: (x−6)2=−4a(y−4)(x-6)^2=-4a(y-4). Note the start (0,0)(0,0) and landing (12,0)(12,0) are symmetric about x=6x=6, consistent with this vertex.

Step 2. Find aa using the start point (0,0)(0,0).

(0−6)2=−4a(0−4)⇒36=16a⇒a=3616=94(0-6)^2=-4a(0-4) \Rightarrow 36=16a \Rightarrow a=\dfrac{36}{16}=\dfrac94.

Step 3. Differentiate to get the slope at any xx.

2(x−6)=−4adydx⇒dydx=−2(x−6)4a=−(x−6)2a2(x-6)=-4a\dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx}=-\dfrac{2(x-6)}{4a}=-\dfrac{(x-6)}{2a}. …

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