$A$ projectile with speed $50 \ m \ s^{-1}$ is thrown at an angle of $60^{\circ}$ with the horizontal. The maximum height that can be reached is,(acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m$)

$A$ 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:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

$A$ shell is fired vertically upwards with a velocity $v_1$ from a trolley moving horizontally with velocity $v_2$. $A$ person on the ground observes the motion of the shell as a parabola,whose horizontal range is ....

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by

$A$ cricketer can throw a ball to a maximum horizontal distance of $100 \; m$. How much high above (in $m$) the ground can the cricketer throw the same ball?

Assertion $(A)$: The range of a projectile is maximum when the angle of projection is $45^{\circ}$.
Reason $(R)$: The range of a projectile depends only on the angle of projection.