A projectile is thrwon into space so as to have the maximum possible horizontal range equal to 400m. Taking the point of projection as the origin, the coordinates of the point where the velocity of the projectile is minimum are:-

Options

A. -400,100
B. -200,100
C. -400,200
D. -200, 200

Answer

B

Explanation

for maximum range `theta=45^(@)`
`because tan theta =(4H)/(R)`
`H=(R)/(4) tan 45^(@)=(400)/(4)=100`
`therefore` required coordinates are `((R)/(2),H)=(200,100)`