The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = 8t - 5t^2 \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The velocity with which the projectile is projected is (in $\text{ m/s}$)

$A$ bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is (in $^{\circ}$)

$A$ ball is thrown at a speed of $20 \,m/s$ at an angle of $30^{\circ}$ with the horizontal. The maximum height reached by the ball is (Use $g=10 \,m/s^2$) (in $\,m$)

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle of $30^{\circ}$ 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:

$A$ heavy particle is projected from a point on the horizontal at an angle $60^\circ$ with the horizontal with a speed of $10 \ m/s$. Then the radius of curvature of its path at the instant of crossing the same horizontal will be ......... $m$.