$A$ ball is projected with a velocity of $10 \ ms^{-1}$ at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be $............... \ ms^{-1}$.

Two projectiles are projected with the same initial velocities at $15^{\circ}$ and $30^{\circ}$ with respect to the horizontal. The ratio of their ranges is $1 : x$. The value of $x$ is:

When a player throws a ball,it reaches the other player in $4 \,s$. If the height of each player is $1.8 \,m$,the maximum height attained by the ball above the ground is (in $\,m$)

$A$ 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 $m$)? $(g = 10 \ m/s^2, \sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$

Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.

$A$ ball is projected with velocity $V_0$ at an angle of elevation $30^o$. Mark the correct statement.