A body is projected horizontally from the surface of the earth (radius `= R`) with a velocity equal to `n` times the escape velocity. Neglect rotational effect of the earth. The maximum height attained by the body from the earth `s` surface is `R//2`. Then, `n` must be

Options

A. `sqrt(0.6)`
B. `(sqrt(3))//2`
C. `sqrt(0.4)`
D. `1//2`

Answer

A

Explanation

`nu_(1) = n nu_(e) = n sqrt(2gR) = n sqrt((2GM)/(R ))`

From mechanical energy conservation, we have
`(1)/(2) m [n sqrt((2GM)/(R ))]^(2) - (GMm)/(R ) = (1)/(2) m nu_(2)^(2) - (GMm)/(R + (R )/(2))` ..(i)
From conservation of momentum about centre of earth
`m [n sqrt((2GM)/(R ))] R = m nu^(2) ((R + (R )/(2))` ..(ii)
Solving these two equation we get,
`n= sqrt(0.6)`