$A$ ball thrown by one player reaches another player in $2 \, s$. The maximum height attained by the ball is ........ $m$. (Take $g = 10 \, m/s^2$)

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:

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g$ is the acceleration due to gravity):

The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$,the range of the projectile (in meters) is:

$A$ 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 $\text{ m}$)