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Find the maximum height attained by a body projected with a speed v=v_(e)//2. Where v_(e )= escape velocity of any object at earth's surface. |
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Answer» Solution :`DeltaKE+DeltaPE=0` `implies-(1)/(2)mv^(2)+GMm((1)/(R )-(1)/(R+h))=0` Put `v=(v_(E ))/(2)=SQRT((GM)/(2R))` `:.h=R//3` |
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