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

If the initial velocity of a projectile is doubled,keeping the angle of projection the same,the maximum height reached by it will:

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g$ is the acceleration due to gravity):

$A$ 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 $g=10 \ m/s^2$)

If a stone is to hit a point which is at a horizontal distance $d$ and at a vertical height $h$ above the point from where the stone is launched,what is the initial speed $u$ if the stone is launched at an angle $\theta$?

$A$ ball is thrown at different angles with the same speed $u$ and from the same point. It has the same range in both cases. If $y_1$ and $y_2$ are the heights attained in the two cases,then $y_1 + y_2$ is equal to: