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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyApplication

JEE Main 26 February 2021, Shift 2, Physics Q9

Question 9 of 90 in this shift, Physics question 9 of 30, Section A.

The trajectory of a projectile in a vertical plane is y=αx−βx2y = \alpha x - \beta x^2, where α\alpha and β\beta are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ\theta and the maximum height attained H are respectively given by :
  1. (1)tan⁡−1β,α22β\tan^{-1}\beta, \frac{\alpha^2}{2\beta}
  2. (2)tan⁡−1(βα),α2β\tan^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^2}{\beta}
  3. (3)tan⁡−1α,α24β\tan^{-1}\alpha, \frac{\alpha^2}{4\beta}Official answer
  4. (4)tan⁡−1α,4α2β\tan^{-1}\alpha, \frac{4\alpha^2}{\beta}

Official answer

Option 3

NTA final key.