$A$ body of mass $1 \, kg$ is projected from the ground at an angle of $30^{\circ}$ with the horizontal on level ground at a speed of $50 \, m/s$. The magnitude of the change in momentum of the body during its entire flight is ....... $kg \cdot m/s$ $(g = 10 \, m/s^2)$.

$A$ ball $A$ is projected vertically upwards with a certain initial speed $u$. Another ball $B$ of the same mass is projected at an angle of $30^{\circ}$ with the vertical with the same initial speed $u$. At the highest point,the ratio of the potential energy of ball $A$ to that of ball $B$ will be: $(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2})$

At what angle should a particle be projected to cover the maximum range (in $^{\circ}$)?

$A$ ball is projected from a building of height $10 \ m$ with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the horizontal. What is the horizontal distance covered by the ball when it reaches the same height of $10 \ m$ again (in $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle of $30^{\circ}$ 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: