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 :-
$\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}$
$\tan ^{-1} \beta, \frac{\alpha^{2}}{2 \beta}$
$\tan ^{-1} \alpha, \frac{4 \alpha^{2}}{\beta}$
$\tan ^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^{2}}{\beta}$
The figure shows a velocity-time graph of a particle moving along a straight line The correct displacement-time graph of the particle is shown as
The figure shows the velocity $(v)$ of a particle plotted against time $(t)$
The equation of a projectile is $y=a x-b x^2$. Its horizontal range is ......
The position of a particle moving in the $xy-$plane at any time $t$ is given by $x = (3{t^2} - 6t)$ metres, $y = ({t^2} - 2t)$ metres. Select the correct statement about the moving particle from the following