$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

$A$ particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

$A$ projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at a height equal to half of the maximum height,the vertical component of the velocity of the projectile is ...........

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

Galileo,in his book $Two \text{ } new \text{ } sciences$,stated that "for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal". Prove this statement.

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$