The trajectory of a projectile in a vertical plane is y = α x -β x2… | JEE Main 2021 (26 Feb, Shift 2)

Q4JEE Main 2021 · 26 FEB · SHIFT 2Medium

Projectile Basics & Trajectory

The trajectory of a projectile in a vertical plane is y = $\alpha$ x $-\beta$ x2, where $\alpha$ and $\beta$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained H are respectively given by:

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Answer: (a)

Compare with $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ : $\tan\theta = \alpha \Rightarrow \theta = \tan^{-1}\alpha$
$\frac{dy}{dx} = 0$ at $x = \frac{\alpha}{2\beta}$
$H = \alpha\cdot\frac{\alpha}{2\beta} - \beta\frac{\alpha^2}{4\beta^2} = \frac{\alpha^2}{4\beta}$
So the answer is (a).

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Projectile Basics & Trajectory question from JEE Main 2021 (26 Feb, Shift 2)

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