The maximum range of a gun on a horizontal plane is $16 \,km$. If $g=10 \,ms^{-2}$,then the muzzle velocity of a shell is

$A$ body is projected from the ground at an angle of $\tan^{-1}(\frac{8}{7})$ with the horizontal. The ratio of the maximum height attained by it to its range is (in $: 7$)

The angle of projection for a projectile to have the same horizontal range and maximum height is

$A$ cricket fielder can throw a cricket ball with a speed $v_{0}$. If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal,find:
$(a)$ The effective angle to the horizontal at which the ball is projected in the air as seen by a spectator.
$(b)$ The time of flight.
$(c)$ The horizontal range from the point of projection at which the ball will land.
$(d)$ The angle $\theta$ at which he should throw the ball to maximize the horizontal range found in $(c)$.
$(e)$ How does $\theta$ for maximum range change if $u > v_{0}$,$u = v_{0}$,and $u < v_{0}$?
$(f)$ How does $\theta$ in $(e)$ compare with that for $u = 0$ (i.e.,$45^{\circ}$)?

The height of the ceiling in an auditorium is $30 \ m$. $A$ ball is thrown with a speed of $30 \ m \ s^{-1}$ from the entrance such that it just moves very near to the ceiling without touching it and then it reaches the ground at the end of the auditorium. Then the length of the auditorium is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is: