A projectile is thrown from ground such that itattains maximum possible horizontal range equal to200m. Taking the point of projection as the origin,the coordinates of the point where the velocity of theprojectile is minimum are; (Neglect air resistance

The velocity of a projectile is minimum at the highest point of its trajectory. This is because the vertical component of the velocity becomes zero at the highest point, while the horizontal component remains constant throughout the flight.

The maximum range (R) of a projectile is given by the formula R = (v^2)/g, where v is the initial velocity and g is the acceleration due to gravity.

Given that the maximum range is 200m, we can rearrange the formula to find the initial velocity: v = sqrt(R*g).

Assuming g is approximately 9.8 m/s^2, the initial velocity v is approximately 44.7 m/s.

The initial velocity can be split into horizontal and vertical components. Since the maximum range is achieved when the launch angle is 45 degrees, these components are equal: v_x = v_y = v/sqrt(2) = approximately 31.6 m/s.

The horizontal component of the velocity remains constant, so the minimum velocity is also 31.6 m/s.

The time to reach the highest point (t) is given by the formula t = v_y/g = approximately 3.22 s.

The horizontal distance travelled in this time (x) is given by the formula x = v_x*t = approximately 101.6 m.

Therefore, the coordinates of the point where the velocity of the projectile is minimum are (101.6, 0).