What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)

Explanation

Let `V_(e)` = escape velocity
`= (1)/(2)m((V_(e))/(3))^(2)`
As `V_(e) = sqrt((2GM)/(R ))`
Therefore, `K.E. = (1)/(2)m (2GM)/(R.9)`
`= (GMm)/(9R)`
Initial `P.E. = (-GMm)/(R )`
Total initial energy = K.E. + P.E.
`= (GMm)/(9R) - (GMm)/(R )`
`= -(8GMm)/(9R)` ...(i)
At maximum height, velocity = 0
so final K.E. = 0
final P.E. `= (-GMm)/(r )`
Total final energy `= (-GMm)/(r )` ...(ii)
From (i) and (ii) we can write
`(-8 GMm)/(9R) = (-GMm)/(r )`
`r = (9)/(8)R`
Therefore maximum height `= r - R = (9)/(8)R - R`
`= (R )/(8)`