A projectile of mass m is fired from the surface of the earth at an angle α = 6 0 ∘ from the vertical. The initial speed v 0 is equal to GM e R e . The maximum height that projectile can rise from the surface of the earth is R e x . Find the value of x ? G = Universal Gravitational constant M e = Mass of earth R e = Radius of earth

Step-by-step solution

By conservation of angular momentum about C at position P and F mv 0 3 2 R e = mv r max ⇒ v = 3 R e v 0 2 r max ...(i) By conservation of mechanical energy between P and F 1 2 mv 0 2 - GM e m R e = 1 2 mv 2 - GM e m r max ...(ii) Put (i) in (ii) 1 2 mv 0 2 - GM e m R e = 1 2 m 3 R 2 v 0 2 4 r max 2 - GM e m r max v 0 = GM e R ⇒ 1 2 m GM e R e - GM e m R e = 3 8 mR e 2 r max 2 GM e R - GM e m r max ⇒ 1 2 R e - 1 R e = 3 R 8 r max 2 - 1 r max ⇒ - 1 2 R e = 3 R e - 8 r max 8 r max 2 ⇒ 4 r max 2 - 8 R e r max + 3 R e 2