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NTA Abhyas JEE Main2020PhysicsMotion in Two DimensionsPractice

The trajectory of a projectile in vertical plane in y = a x - b x 2 , where a and b are constant and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal is:

Options

  1. Aa 2 4 b , tan - 1 ( b )
  2. Ba 2 b , tan - 1 ( 2 b )
  3. Ca 2 4 b , tan - 1 ( a )
  4. D2 a 2 b , tan - 1 ( a )

Correct answer

C. a 2 4 b , tan - 1 ( a )

Step-by-step solution

Let the angle of projection be θ y = ax - bx 2 To get maxima, dy dx = 0 dy dx = a - b . 2x = 0 ⇒ x = a 2b y max = a 2 2 b - ba 2 4 b 2 = a 2 4b Angle of projections can be obtained by slope dy dx at x = 0 ⇒ Slope at x = 0 is a ⇒ tan θ = a ⇒ θ = tan -1 a

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