The equation of motion of a projectile is $y = Ax - Bx^2$ where $A$ and $B$ are constants of motion. The horizontal range of the projectile is

If the time of flight of a bullet over a horizontal range $R$ is $T$,then the angle of projection with the horizontal is ......

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

$A$ particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ shell is fired at an angle of $30^{\circ}$ to the horizontal with a velocity of $196 \,m/s$. What is the time of flight (in $\,s$)? (Take $g = 9.8 \,m/s^2$)

The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$, respectively. Initial velocity of the projectile in $m/s$ is