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

The correct option is (R/2, R/4).

To find the coordinates of the point where the speed of the projectile is minimum, let us analyze the motion step-by-step:

Step 1: Determine the angle of projection for maximum horizontal range
The horizontal range $R$ of a projectile launched with an initial velocity $u$ at an angle $\theta$ to the horizontal is given by the formula:

$R=\frac{u^{2}sin\left(2\theta \right)}{g}$

where $g$ is the acceleration due to gravity. For the range to be maximum, the term $sin\left(2\theta \right)$ must be equal to its maximum value, which is 1:

$sin\left(2\theta \right)=1\Rightarrow 2\theta ={90}^{\circ }\Rightarrow \theta ={45}^{\circ }$

Substituting this back, the maximum range $R$ is:

$R=\frac{u^2}{g}$

Step 2: Identify the point of minimum speed
At any point during the flight, the velocity vector of the projectile has two components:
1. A constant horizontal component: $v_x=ucos\theta$
2. A variable vertical component: $v_y=usin\theta -gt$

The speed of the particle at any instant is:

$v=\sqrt{v_x^2+v_y^2}$

Since $v_x$ is constant, the speed $v$ is minimum when the vertical component of velocity becomes zero ($v_y=0$). This occurs at the highest point (peak) of the projectile's trajectory.

Step 3: Calculate the coordinates of the highest point
Taking the point of projection as the origin $(0,0)$, the coordinates of the highest point are given by $(x,y)$, where:
- $x$ is half of the horizontal range:

$x=\frac{R}{2}$

- $y$ is the maximum height $H$ reached by the projectile:

$y=H=\frac{u^2{sin}^2\theta }{2g}$

Since $\theta ={45}^{\circ }$, we have:

${sin}^2\left({45}^{\circ }\right)={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$

Substituting this value into the height formula:

$y=\frac{u^2}{2g}\cdot \frac{1}{2}=\frac{u^2}{4g}$

Using our expression for the maximum range $R=\frac{u^2}{g}$:

$y=\frac{R}{4}$

Therefore, the coordinates of the point where the speed of the projectile is minimum are:

$(x,y)=(\frac{R}{2},\frac{R}{4})$