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Q.

Let the escape speed of an object on the earth’s surface be  V0. The object is projected out with speed  5V0 .  The speed of the object far away from the earth will be 

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a

23V0

b

26V0

c

32V0

d

4V0

answer is A.

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Detailed Solution

VSpace=Vp2−Ve2

Vp = projected velocity 

Vspace=5V02−V02=24V02=26V0

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