The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

If the time of flight of a projectile is $10 \ s$,and its range is $500 \ m$,then the maximum height attained by it will be ......... $m$.

$A$ gun is aimed at a target in the line of its barrel. The target is released and allowed to fall under gravity at the same instant the gun is fired. The bullet will

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is:

For a projectile, if $\alpha$ is the angle of projection, $R$ is the range, $h$ is the maximum height, and $T$ is the time of flight, then:

$A$ 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