The trajectory a projectile in a vertical plane

Question:

The trajectory a projectile in a vertical plane is $y=\alpha x-\beta x^{2}$, where $\alpha$ and $\beta$ are constants and $x \backslash y$ are respectively the horizontal and vertical distance 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. (1) $\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}$

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

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

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

Correct Option: 1

Solution:

(1)

Given :

$y=\alpha x-\beta x^{2} \quad \ldots(1)$

for maximum height, we should find out maximum value of y from equation (1) so, for maximum value of y

$\frac{d y}{d x}=0 \Rightarrow \alpha-2 \beta x=0$

$x=\frac{\alpha}{2 \beta} \quad \ldots(2)$

Now, put value of $x$ from equation (2) in quation (1) $y=\alpha\left(\frac{\alpha}{2 \beta}\right)-\beta\left(\frac{\alpha^{2}}{4 \beta^{2}}\right)$

$\Rightarrow\left(\frac{\alpha^{2}}{2 \beta}\right)-\left(\frac{\alpha^{2}}{4 \beta}\right) \Rightarrow \frac{\alpha^{2}}{4 \beta}$

So, $\mathrm{H}_{\max }=\frac{\alpha^{2}}{4 \beta} \quad \ldots(3)$

As we know maximum height $\mathrm{H}_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g} \quad \ldots$ (4)

from (3) and (4) $\mathrm{u}^{2}=\left(\frac{\alpha^{2}}{4 \beta}\right)\left(\frac{2 g}{\sin ^{2} \theta}\right)$

and range $(\mathrm{R})=2 \mathrm{x}=\frac{\mathrm{u}^{2} \times 2 \sin \theta \cos \theta}{\mathrm{g}}$

$2\left(\frac{\alpha}{2 \beta}\right)=\frac{\left(\frac{\alpha^{2}}{4 \beta}\right)\left(\frac{2 g}{\sin ^{2} \theta}\right)_{g}^{\times 2 \sin \theta \cos \theta}}{g}$

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