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))

A projectile is launched from the surface of the earth with a very hig

1.	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))
Answer» `(U "cos" theta)/(2)` `( u "SIN" theta)/(2)` `SQRT((GM)/(2R))` `sqrt((GM)/(3R))` Answer :B

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