: A projectile is projected with velocity $k{v_e}$ in vertically upward direction from the ground into the space ( ${v_e}$ is escape velocity and\[k < 1\] ). If air resistance is considered to be negligible, then the maximum height from the centre of earth till which it can go will be (R= radius of earth).
(A) $\dfrac{R}{{{k^2} + 1}}$
(B) $\dfrac{R}{{{k^2} - 1}}$
(C) $\dfrac{R}{{1 - {k^2}}}$
(D) $\dfrac{R}{{k + 1}}$

Hint: Calculate the potential final and initial gravitational potential energy to and subtract the initial gravitational potential energy from final gravitational potential energy to get the gravitational kinetic energy and equate it with the projectile’s kinetic energy to get the maximum height travelled by the projectile.

Complete step by step answer
Gravitational force is an invisible force that pulls any two objects towards each other. Newton’s law of gravity states that every particle in the universe attracts every other particle with a force is equal to the product of two objects of masses divided by the square of their distance. The closer the objects are, the stronger the gravitational force.
Gravitational potential energy: the potential energy stored in an object is due to the gravitational force acting on the object. Every object in gravitational field has gravitational potential energy
The gravitational potential energy is calculated as follow
${U_g} = - \dfrac{{GMm}}{R}$
Where,
$M$ is the mass of the object 1
$m$ is the mass of the object 2
$R$ is the centre to centre distance between the object 1 and 2
$G$ is the gravitational constant
${U_g}$ is the gravitational potential energy
Given,
Potential energy at surface is ${U_i} = - \dfrac{{GMm}}{R}$
Potential energy at height h from the centre is final Potential energy is ${U_f} = - \dfrac{{GMm}}{h}$
$M$ is the mass of the earth
$m$ is the mass of the projectile
$R$ is the distance
$G$ is the gravitational constant
${U_i}$ is the initial gravitational potential energy at the surface of the earth
${U_f}$is the final potential energy at the centre at the height h
Difference in potential energy is ${U_f} - {U_i} = - \dfrac{{GMm}}{R} - ( - \dfrac{{GMm}}{h})$
$ \Rightarrow {U_f} - {U_i} = GMm( - \dfrac{1}{R} + \dfrac{1}{h})$
$ \Rightarrow {U_f} - {U_i} = GMm(\dfrac{{R - h}}{{rh}})$
$ \Rightarrow {U_f} - {U_i} = \dfrac{{GMm}}{R}(\dfrac{{R - h}}{h})$
$ \Rightarrow {U_f} - {U_i} = \dfrac{{GMm}}{R}(\dfrac{R}{h} - 1)$
$ \Rightarrow K.E = \dfrac{{GMm}}{R}(\dfrac{R}{h} - 1){\text{ }} \to 1$
This difference in potential energy is nothing but the kinetic energy supplied at the time of projection.
Kinetic energy is $K.E = \dfrac{1}{2}m{v^2}$
Velocity is of the projectile projected upwards is $k \times {v_e}$ , where ${v_e}$ is the escape velocity
We know that escape velocity is ${v_{esc}} = \sqrt {\dfrac{{2GM}}{R}} $
Then, Velocity is of the projectile projected upwards is $\left( {k\sqrt {\dfrac{{2GM}}{R}} } \right)$
$G$ is the gravitational constant
$M$ is the mass of the earth
$R$ is the radius of earth
Substitute all the above values in the formula.
$K.E = \dfrac{1}{2}m{v^2}$
$K.E = \dfrac{1}{2}m{\left( {k\sqrt {\dfrac{{2GM}}{R}} } \right)^2}$
$K.E = \dfrac{1}{2}m \times {k^2} \times {\left( {\sqrt {\dfrac{{2GM}}{R}} } \right)^2}$
$K.E = {k^2}\dfrac{{GMm}}{R}{\text{ }} \to {\text{2}}$
Equating the 1 and 2 we get
$\dfrac{{GMm}}{R}(\dfrac{R}{h} - 1){\text{ = }}{k^2}\dfrac{{GMm}}{R}$
$\dfrac{R}{h} - 1{\text{ = }}{k^2}$
$\dfrac{R}{h}{\text{ = }}{k^2} + 1$
$h = \dfrac{R}{{{k^2} + 1}}$

Hence the correct answer is option (A) $\dfrac{R}{{{k^2} + 1}}$

Note: The gravitational potential energy is a negative value because work is done by the gravitational field in bringing a mass from infinity if it is taken away from the earth.