The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.

$A$ particle of mass $m$ is projected with a velocity $V$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ particle is projected with velocity $2 \sqrt{gh}$ at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2h$ from each other. The time taken by the particle to travel between these two walls is

$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........

$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)

$A$ cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is