A projectile of mass $m$ is fired from the surface of the earth at an angle $\alpha =60^{\circ }$ from the vertical. The initial speed $v_0$ is equal to $\sqrt{\frac{{\mathit{\text{GM}}}_{\text{e}}}{{\mathit{\text{R}}}_{\text{e}}}}$. The maximum height that projectile can rise from surface of earth is $\frac{{\mathit{\text{R}}}_{\text{e}}}{\mathit{\text{x}}}$. Find the value of $x$ ?

$G$ = Universal Gravitational constant
$M_{\mathit{e}}$ = Mass of earth
$R_e$ = Radius of earth

• Correct answer
  A

  2
• B

  4

• C

  6

• D

  8

Solution

By conservation of angular momentum about C at position P and F

${\text{mv}}_0\frac{\sqrt{3}}{2}{\text{R}}_{\text{e}}=\text{mv}{r}_{\text{max}}$

$\Rightarrow \text{v}=\frac{\sqrt{3}{\text{R}}_{\text{e}}{\text{v}}_0}{2r_{\text{max}}}$           ...(i)

By conservation of mechanical energy between P and F

$\frac{1}{2}{\text{mv}}_0^2-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{R}}_{\text{e}}}=\frac{1}{2}{\text{mv}}^2-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{r}}_{\text{max}}}$            ...(ii)

Put (i) in (ii)

$\frac{1}{2}{\text{mv}}_0^2-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{R}}_{\text{e}}}=\frac{1}{2}\text{m}\frac{3{\text{R}}^2{\text{v}}_0^2}{4r_{\text{max}}^2}-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{r}}_{\text{max}}}$

${\text{v}}_0=\sqrt{\frac{{\text{GM}}_{\text{e}}}{\text{R}}}$

$\Rightarrow \frac{1}{2}\text{m}\frac{{\text{GM}}_{\text{e}}}{{\text{R}}_{\text{e}}}-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{R}}_{\text{e}}}=\frac{3}{8}\frac{{\text{mR}}_{\text{e}}^2}{{\text{r}}_{\text{max}}^2}\frac{{\text{GM}}_{\text{e}}}{\text{R}}-\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{r}}_{\text{max}}}$

$\Rightarrow \frac{1}{2{\text{R}}_{\text{e}}}-\frac{1}{{\text{R}}_{\text{e}}}=\frac{3\text{R}}{8r_{\text{max}}^2}-\frac{1}{{\text{r}}_{\text{max}}}$

$\Rightarrow \frac{-1}{2{\text{R}}_{\text{e}}}=\frac{3{\text{R}}_{\text{e}}-8r_{\text{max}}}{8r_{\text{max}}^2}$

$\Rightarrow 4{\text{r}}_{\text{max}}^2-8{\text{R}}_{\text{e}}{r}_{\text{max}}+3{\text{R}}_{\text{e}}^2=0$

Solving the quadratic in r_max

we get

${\text{r}}_{\text{max}}=\frac{8{\text{R}}_{\text{e}}\pm \sqrt{16}{\text{R}}_{\text{e}}}{8}=\frac{8{\text{R}}_{\text{e}}\pm 4{\text{R}}_{\text{e}}}{8}$

$\Rightarrow {\text{r}}_{\text{max}}=\frac{8{\text{R}}_{\text{e}}+4{\text{R}}_{\text{e}}}{8}=\frac{12{\text{R}}_{\text{e}}}{8}=\frac{3{\text{R}}_{\text{e}}}{2}$

$\therefore {\text{h}}_{\text{max}}={\text{r}}_{\text{max}}-{\text{R}}_{\text{e}}=\frac{3{\text{R}}_{\text{e}}}{2}-{\text{R}}_{\text{e}}=\frac{{\text{R}}_{\text{e}}}{2}$

$\therefore \text{x}=2$

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