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Question 83 of 126

Q.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.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 10mImportance★★★★★
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Model the path as a parabola with vertex at the point of maximum height, use the fact that the trajectory's coefficient of xx equals tan⁡θ\tan\theta, and read this off the parabola's equation.

  1. The rocket starts at the origin (0,0)(0,0), reaches maximum height H=4H=4 m at horizontal distance 66 m (the vertex), and lands at (12,0)(12,0).
  2. Since the vertex is at x=6=122x=6=\dfrac{12}{2}, this confirms the maximum-height point is the midpoint of the range R=12R=12 m, consistent with a symmetric parabolic trajectory.
  3. Let the trajectory be the parabola with vertex (6,4)(6,4): y=a(x−6)2+4y = a(x-6)^2+4.
  4. Since the path starts at the origin, y(0)=0y(0)=0: 0=a(0−6)2+4=36a+4⇒a=−436=−190 = a(0-6)^2+4 = 36a+4 \Rightarrow a=-\dfrac{4}{36}=-\dfrac19.
  5. So y=−19(x−6)2+4y = -\dfrac19(x-6)^2+4. Expanding: y=−19(x2−12x+36)+4=−x29+12x9−4+4=43x−x29y = -\dfrac19(x^2-12x+36)+4 = -\dfrac{x^2}{9}+\dfrac{12x}{9}-4+4 = \dfrac{4}{3}x-\dfrac{x^2}{9}.
  6. The standard projectile trajectory equation is y=xtan⁡θ−g2u2cos⁡2θx2y = x\tan\theta - \dfrac{g}{2u^2\cos^2\theta}x^2, whose coefficient of xx is tan⁡θ\tan\theta. …
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