$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

$A$ projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

The path of a projectile is given by the equation $y = ax - bx^2$,where $a$ and $b$ are constants,and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection,respectively. The maximum height attained by the projectile and the angle of projection are respectively:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: When a body is projected at an angle $45^{\circ}$,its range is maximum.
Reason $R$: For maximum range,the value of $\sin 2\theta$ should be equal to one.
In the light of the above statements,choose the correct answer from the options given below:

$A$ 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?

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.