An object is thrown directly away from the surface of the earth with an initial speed v . The object reaches upto a height of 4 5 R_E from earth's surface, where R_E is radius of the earth. If the escape velocity of the object is v_E then the value of v v_E is

Step-by-step solution

We know that, maximum height attained by a projectile projected with velocity v . h= v^2 R_E 2 g R_E-v^2 But given, h= 4 5 R_E aligned & & 4 5 R_E & = v^2 R_E 2 g R_E-v^2 & & 4 5 & = v^2 2 g R_E-v^2 & & 5 v^2 & =8 g R_E-4 v^2 & & 9 v^2 & =8 g R_E & & v & = 8 9 g R_E = 2 3 2 g R_E & & v & = 2 3 , v_E [ v_E= 2 g R_E ] & & v v_E & = 2 3 aligned