A projectile is launched from the surface of the earth with a very high speed u at an angle `theta` with vertical . What is its velocity when it is at the farthest distance from the earth surface . Given that the maximum height reached when it is launched vertically from the earth with a velocity v = `sqrt((GM)/(R))`

Options

A. `(u "cos" theta)/(2)`
B. `( u "sin" theta)/(2)`
C. `sqrt((GM)/(2R))`
D. `sqrt((GM)/(3R))`

Answer

B