The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)

$A$ fighter plane, flying horizontally with a speed of $360 \text{ km/h}$ at an altitude of $500 \text{ m}$, drops a bomb for a target straight ahead of it on the ground. At what approximate distance ahead of the target should the bomb be dropped? Assume that acceleration due to gravity $g = 10 \text{ m/s}^2$. Neglect air drag.

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal is given as $y = x - \frac{gx^2}{2}$. The initial velocity of the ball is $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings $100 \; m$ apart and of the same height of $200 \; m$ with the same velocity of $25 \; m/s$. When and where will the two bullets collide? $(g = 10 \; m/s^2)$

$A$ body when projected at an angle $\theta$ with the horizontal reaches a maximum height $H$. The time of flight of the body will be ($g=$ acceleration due to gravity).

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is