The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$, respectively. Initial velocity of the projectile in $m/s$ is

$A$ bomb is dropped by an airplane flying horizontally with a velocity of $200 \text{ km/hr}$ at a height of $980 \text{ m}$. At the time of dropping the bomb,the horizontal distance of the airplane from the target on the ground to hit it directly is (given $g = 9.8 \text{ m/s}^2$):

$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)

The maximum horizontal range of a projectile is $400\;m$. What is the maximum height attained by it (in $;m$)?

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by