$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $= g$)

The maximum range of a gun on horizontal terrain is $16 \, km$. If $g = 10 \, m/s^2$,what must be the muzzle velocity of the shell in $m/s$?

$A$ 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

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147 \ m/s$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is ......... $s$.

$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

$A$ particle is projected with a velocity of $30\,m/s$ at an angle of $\theta_0 = \tan^{-1}(3/4)$. After $1\,s$,the particle is moving at an angle $\theta$ to the horizontal. What is the value of $\tan\theta$? (Take $g = 10\,m/s^2$)