$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

In the climax of a movie,the hero jumps from a helicopter and the villain chasing the hero also jumps at the same time from the same level. After some time,when they are at the same horizontal level,the villain fires a bullet horizontally towards the hero. Both the hero and the villain are falling with a constant downward acceleration of $2\ m/s^2$ due to their parachutes. Assuming the hero is within the range of the bullet and air resistance on the bullet is negligible,which of the following is correct?

$A$ projectile is launched from the ground, such that it hits a target on the ground which is $90 \,m$ away. The minimum velocity of the projectile to hit the target is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-1}$)

$A$ body of mass $M$ thrown horizontally with velocity $v$ from the top of a tower of height $H$ touches the ground at a distance of $100 \ m$ from the foot of the tower. $A$ body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of a tower of height $4H$ will touch the ground at a distance of ........ $m$.

$A$ body is projected at an angle of $30^{\circ}$ with the horizontal with momentum $p$. At its highest point,the magnitude of the momentum is

$A$ bullet is fired at time $t=0$ with velocity $20 \ m/s$ and at an initial angle of $30^{\circ}$ with the horizontal. The angle between the displacement vector and the horizontal after time $0.1 \ s$ is (Assume $g=10 \ m/s^2$)