$A$ 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

$A$ projectile is given an initial velocity of $\hat{i}+2 \hat{j} \,ms^{-1}$. The Cartesian equation of its path is ($x$ and $y$ are in metres and $g=10 \,ms^{-2}$)

$A$ particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in a horizontal direction with a speed of $5 \; m/s$ from the same height. The correct statement is:

$A$ cannon on a level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired with a muzzle velocity $v_0$ towards a vertical cliff a distance $D$ away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

$A$ body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $XY$-plane passes through a point at $(4, 3) \ m$. Its horizontal range is (in $m$)

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $= g$)