A particle is projected with velocity $k{v}_e$ in vertically upward direction from the ground into the space ( $v_e$ is escape velocity & $k<1$ ) then the maximum height from the centre of earth to which it can go, will be

• A

  $\frac{R}{k^2+1}$
• B

  $\frac{R}{k^2–1}$
• Correct answer
  C

  $\frac{R}{1–k^2}$
• D

  $\frac{R}{k+1}$

Hence, the correct answer is option: 'C'.