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)(1) (100, 70) m(2) (200, 50) m(3) (100, 50) m(4) (100, 100) m

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 height (H) of a projectile launched at an angle θ with an initial velocity u is given by the formula H = u²sin²θ/2g. For maximum range, the launch angle θ is 45 degrees.

The range (R) of a projectile is given by the formula R = u²sin2θ/g. For maximum range and a launch angle of 45 degrees, this simplifies to R = u²/g.

Given that the maximum range is 200m, we can set R = 200m and solve for u:

200 = u²/g u = sqrt(200g)

Substituting this back into the formula for maximum height:

H = (sqrt(200g))²sin²45/2g H = 100m

The x-coordinate at the highest point is half the total range, or 100m.

So, the coordinates of the point where the velocity of the projectile is minimum are (100, 100) m. Therefore, the correct answer is (4) (100, 100) m.