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:
Correct Option: 1
(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 $(\mathrm{R})=2 \mathrm{x}=\frac{\mathrm{u}^{2} \times 2 \sin \theta \cos \theta}{\mathrm{g}}$
$2\left(\frac{\alpha}{2 \beta}\right)=\frac{\left(\frac{\alpha^{2}}{4 \beta}\right)\left(\frac{2 g}{\sin ^{2} \theta}\right)_{g}^{\times 2 \sin \theta \cos \theta}}{g}$
$\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1}(\alpha)$
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