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3-2.Motion in Plane
hard
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$\tan ^{-1} \alpha, \frac{4 \alpha^{2}}{\beta}$
D$\tan ^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^{2}}{\beta}$
(JEE MAIN-2021)
Solution
$y =\alpha x -\beta x ^{2}$
comparing with trajectory equation
$y = x \tan \theta-\frac{1}{2} \frac{ gx ^{2}}{ u ^{2} \cos ^{2} \theta}$
$\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1} \alpha$
$\beta=\frac{1}{2} \frac{ g }{ u ^{2} \cos ^{2} \theta}$
$u ^{2}=\frac{ g }{2 \beta \cos ^{2} \theta}$
Maximum height : $H$
$H =\frac{ u ^{2} \sin ^{2} \theta}{2 g }=\frac{ g }{2 \beta \cos ^{2} \theta} \frac{\sin ^{2} \theta}{2 g }$
$H =\frac{\tan ^{2} \theta}{4 \beta}=\frac{\alpha^{2}}{4 \beta}$
comparing with trajectory equation
$y = x \tan \theta-\frac{1}{2} \frac{ gx ^{2}}{ u ^{2} \cos ^{2} \theta}$
$\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1} \alpha$
$\beta=\frac{1}{2} \frac{ g }{ u ^{2} \cos ^{2} \theta}$
$u ^{2}=\frac{ g }{2 \beta \cos ^{2} \theta}$
Maximum height : $H$
$H =\frac{ u ^{2} \sin ^{2} \theta}{2 g }=\frac{ g }{2 \beta \cos ^{2} \theta} \frac{\sin ^{2} \theta}{2 g }$
$H =\frac{\tan ^{2} \theta}{4 \beta}=\frac{\alpha^{2}}{4 \beta}$
Standard 11
Physics
