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 $m/s$?

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle $45^o$ with the horizontal. The magnitude of the torque due to the weight of the projectile,when the particle is at its maximum height,about the point of projection is:

Two paper screens $A$ and $B$ are separated by $150 \ m$. $A$ bullet pierces $A$ and then $B$. The hole in $B$ is $15 \ cm$ below the hole in $A$. If the bullet is traveling horizontally at the time of hitting $A$,then the velocity of the bullet at $A$ is: $(g=10 \ m \ s^{-2})$

$A$ body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up,the velocity of the mass after $t$ seconds will be:

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)