$A$ 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

$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

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:

$A$ stone is projected horizontally with a speed $10 \ m/s$ from an $80 \ m$ high building. The distance of the target on the ground from the foot of the building is $.... \ m$ $(g = 10 \ m/s^2)$.

At the highest point of the path of a projectile,its

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)