The maximum height reached by a projectile is $64 \,m$. If the initial velocity is halved, the new maximum height of the projectile is . . . . . . $m$.

$A$ bomb of mass $m$ is projected from the ground with speed $v$ at an angle $\theta$ with the horizontal. At the maximum height from the ground,it explodes into two fragments of equal mass. If one fragment comes to rest immediately after the explosion,then the horizontal range of the centre of mass is ..............

$A$ ball rolls from the top of a stairway with a horizontal velocity $u \; m/s$. If the steps are $h \; m$ high and $b \; m$ wide,the ball will hit the edge of the $n^{th}$ step,if $n=$

$A$ particle of mass $m$ is projected with a velocity $V$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

$A$ boy weighing $50 \ kg$ finished a long jump at a distance of $8 \ m$. Considering that he moved along a parabolic path and his angle of jump is $45^{\circ}$,his initial $K.E.$ is (in $J$)