JEE PYQ: Motion In Two Dimensions Question 7
Question 7 - 2021 (26 Feb Shift 2)
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:
(a) $\tan^{-1}\alpha$, $\frac{\alpha^2}{4\beta}$ (b) $\tan^{-1}\beta$, $\frac{\alpha^2}{2\beta}$ (c) $\tan^{-1}\left(\frac{\beta}{\alpha}\right)$, $\frac{\alpha^2}{2\beta}$ (d) $\tan^{-1}\alpha$, $\frac{\alpha^2}{2\beta}$
Show Answer
Answer: (a)
Solution
$dy/dx = 0 \Rightarrow x = \alpha/(2\beta)$. $H = \alpha^2/(4\beta)$. $\tan\theta = \alpha$, so $\theta = \tan^{-1}\alpha$.
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