$A$ bullet is fired with a velocity $u$ making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.

The path of a projectile is given by the equation $y = ax - bx^2$,where $a$ and $b$ are constants,and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection,respectively. The maximum height attained by the projectile and the angle of projection are respectively:

$A$ ball is thrown from the ground to clear a wall $3 \, m$ high at a distance of $6 \, m$ and falls $18 \, m$ away from the wall. The angle of projection of the ball is:

$A$ projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at a height equal to half of the maximum height,the vertical component of the velocity of the projectile is ...........

$A$ gun and a target are at the same horizontal level separated by a distance of $600 \,m$. The bullet is fired from the gun with a velocity of $500 \,ms^{-1}$. In order to hit the target, the gun should be aimed to a height $h$ above the target. The value of $h$ is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,m$)