A body is thrown from the surface of the earth with velocity u m/s. The maximum height in metre above the surface of the earth up to which it will reach is (where, R = radish of earth, g=acceleration due to gravity)

Options

A. `(u^(2)R)/(2g(R-u^(2)))`
B. `(2u^(2)R)/(gR-u^(2))`
C. `(u^(2)R^(2))/(2gR^(2)-u^(2))`
D. `(u^(2)R)/(gR-u^(2))`

Answer

A

Explanation

Given, velocity of the body = u m/s
According to the question,
`(-GMm)/(R)+(1)/(2)m u^(2)=0+((-GMm)/(R+h))`
`(GM)/(R+h)=(GM)/(R)=(u^(2))/(2)`
`rArr" "(GM)/(R+h)=(2GM-Ru^(2))/(2R)`
`rArr" "(R+h)/(GM)=(2R)/(2Gm-Ru^(2))`
`rArr" "h=(2gMR)/(2(GM-Ru^(2))-R`
`rArr" "h=(2GMR-2GMR+R^(2)u^(2))/(2GM-Ru^(2))`
`rArr" "h(R^(2)u^(2))/(2GM-Ru^(2))` . . .(i)
But `g=(GM)/(R^(2))`
`rArr" "GM=gR^(2)`
Substituting the value of Eq. (ii) in Eq. (i), we get
`rArr" "h=(R^(2)u^(2))/(2gR^(2)-Ru^(2))`
`rArr" "h=(Ru^(2))/(2gR-u^(2))`