$A$ boy throws a ball with a velocity $u$ at an angle $\theta$ with the horizontal. At the same instant,he starts running with a uniform velocity to catch the ball before it hits the ground. To achieve this,he should run with a velocity of

The angular momentum of a projectile projected at an angle $\theta$ with the horizontal with speed $u$ about the point of projection when it is at the highest point of its trajectory is

If the initial velocity of a projectile is doubled,keeping the angle of projection the same,the maximum height reached by it will:

$A$ projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as the origin,the coordinates of the point where the speed of the particle is minimum are:

$A$ body is projected horizontally from a tower with a velocity of $18 \, ms^{-1}$. It strikes the ground at an angle of $45^o$. The vertical component of the velocity when it strikes the ground is ........ $ms^{-1}$.

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: