$A$ ball is projected horizontally with a velocity of $5 \text{ m/s}$ from the top of a building $19.6 \text{ m}$ high. How long will the ball take to hit the ground?

$A$ ball is projected with velocity $V_0$ at an angle of elevation $30^o$. Mark the correct statement.

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 $\text{ m/s}$)

$A$ bomb is dropped on an enemy post on the ground by an aeroplane flying horizontally with a velocity of $60 \ km/h$ at a height of $490 \ m$. At the time of dropping the bomb,the horizontal distance of the aeroplane from the enemy post so that the bomb hits the target is

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is: (projection speed = $u$,angle of projection from horizontal = $\theta$)