A projectile is thrown so as to have a maximum possible horizontal range of $400\text{m}$. Taking the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum is

The correct option is B $(200,100)$
$R=\frac{u^2\sin 2\theta }{g}$
For $R$ to be maximum, $\sin 2\theta$ should be maximum
$\Rightarrow \sin 2\theta =1\Rightarrow 2\theta ={90}^{\circ }\Rightarrow \theta ={45}^{\circ }$
According to question
$R=\frac{u^2}{g}=400$
We know that velocity is minimum at the highest point of projectile (i.e. $H$)
By using the relation,
$R\tan \theta =4H$
$400\tan {45}^{\circ }=4H$
$\Rightarrow H=100$

$\therefore$ Coordinates of the point where velocity is minimum is $(200,100)$
Hence, the correct answer is option (b).
Note: The vertical line passing through highest point divides the Range into two equal parts.