$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)

$A$ particle $P$ is projected with velocity $u_1$ at an angle of $30^\circ$ with the horizontal. Another particle $Q$ is projected vertically upwards with velocity $u_2$ from the point directly below the maximum height of $P$. The condition for the collision of both particles is:

$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The velocity of projection is........ $ms^{-1}$.

$A$ particle is projected at an angle of $45^{\circ}$ with the horizontal with an initial kinetic energy $K$. What will be its kinetic energy at the highest point?

$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)