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 :

\tan ^ - 1\alpha ,\alpha ^2 \over 4\beta

\tan ^ - 1\alpha ,4\alpha ^2 \over \beta

\tan ^ - 1( \beta \over \alpha ),\alpha ^2 \over \beta

\tan ^ - 1\beta ,\alpha ^2 \over 2\beta

Motion in A Plane PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q59 with Solution

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Correct answer: A

For a projectile projected with speed $u$ at angle $\theta$ , its trajectory is

y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}

The given trajectory is

y = \alpha x - \beta x^2

Comparing coefficients:

\tan\theta = \alpha

So,

\theta = \tan^{-1}(\alpha)

The trajectory is

y = \alpha x - \beta x^2

This is a downward opening parabola. The maximum height is the maximum value of $y$ .

For

y = -\beta x^2 + \alpha x

vertex occurs at

x = -\frac{b}{2a}

Here, $a=-\beta$ and $b=\alpha$ , so

x_{\text{vertex}} = -\frac{\alpha}{2(-\beta)} = \frac{\alpha}{2\beta}

Substitute into $y$ :

H = \alpha\left(\frac{\alpha}{2\beta}\right) - \beta\left(\frac{\alpha}{2\beta}\right)^2

H = \frac{\alpha^2}{2\beta} - \beta\cdot \frac{\alpha^2}{4\beta^2}

H = \frac{\alpha^2}{2\beta} - \frac{\alpha^2}{4\beta}

H = \frac{\alpha^2}{4\beta}

Thus,

\theta = \tan^{-1}(\alpha), \qquad H = \frac{\alpha^2}{4\beta}

This matches Option A.

Stored correct answer: A

Derived answer: A

So they agree.