# The trajectory of a projectile in a vertical plane is

Question:

The trajectory of a projectile in a vertical plane is $\mathrm{y}=\alpha \mathrm{x}-\beta \mathrm{x}^{2}$, 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 $\mathrm{H}$ are respectively given by :-

1. $\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}$

2. $\tan ^{-1} \beta, \frac{\alpha^{2}}{2 \beta}$

3. $\tan ^{-1} \alpha, \frac{4 \alpha^{2}}{\beta}$

4. $\tan ^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^{2}}{\beta}$

Correct Option: 1

Solution:

$y=\alpha x-\beta x^{2}$

comparing with trajectory equation

$y=x \tan \theta-\frac{1}{2} \frac{g x^{2}}{u^{2} \cos ^{2} \theta}$

$\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1} \alpha$

$\beta=\frac{1}{2} \frac{\mathrm{g}}{\mathrm{u}^{2} \cos ^{2} \theta}$

$\mathrm{u}^{2}=\frac{\mathrm{g}}{2 \beta \cos ^{2} \theta}$

Maximum height : $\mathrm{H}$

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{g}}{2 \beta \cos ^{2} \theta} \frac{\sin ^{2} \theta}{2 \mathrm{~g}}$

$\mathrm{H}=\frac{\tan ^{2} \theta}{4 \beta}=\frac{\alpha^{2}}{4 \beta}$