Q 11-03-135JEE MainJEE Main 2021 (26 Feb, Shift 2)Easy
The trajectory of a projectile in a vertical plane is $y = \alpha x - \beta x^2$, 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
Answer: (D) $\tan^{-1}\alpha,\ \frac{\alpha^2}{4\beta}$
At the point of projection, $\dfrac{dy}{dx} = \alpha - 2\beta x = \alpha$ at $x = 0$, so $\tan\theta = \alpha$.
Maximum height where $\dfrac{dy}{dx} = 0$: $x = \dfrac{\alpha}{2\beta}$.
$$H = \alpha\cdot\frac{\alpha}{2\beta} - \beta\cdot\frac{\alpha^2}{4\beta^2} = \frac{\alpha^2}{4\beta}$$
Solution by Sreeraj P, M.Sc Physics