Question

A projectile is fired from the surface of Earth at speed of $\sqrt{\frac{GM}{R}}$ at an angle of ${30}^{\circ }$ from vertical then,

• A. minimum speed attained by projectile is $\sqrt{\frac{GM}{4R}}$.
• B. maximum height attained from the surface of Earth is $\frac{\sqrt{3}R}{2}$.
• C. path of projectile is parabolic.
• D. angle at which projectile hits Earth is ${30}^{\circ }$ from vertical.

Answer 1

Answer: (B), (D)

The correct options are
B maximum height attained from the surface of Earth is $\frac{\sqrt{3}R}{2}$.
D angle at which projectile hits Earth is ${30}^{\circ }$ from vertical.


By conservation of angular momentum about C,

$m{V}_0\sin {30}^{\circ }R=mVr$ $[V_0=\sqrt{\frac{GM}{R}}]$

$\frac{V_{0}R}{2}=Vr\dots \dots \dots \left(1\right)$

By Conservation of Energy,

$\Rightarrow \frac{-GM\text{/}m}{R}+\frac{1}{2}\text{/}m{V}_0^2=-\frac{GM\text{/}m}{r}+\frac{1}{2}\text{/}m{V}^2\dots \dots \dots \left(2\right)$

$\frac{-GM}{R}+\frac{1}{2}\frac{GM}{R}=\frac{V^2}{2}-\frac{GM}{R}\Rightarrow \frac{V^2}{2}=\frac{GM}{r}-\frac{GM}{2R}$

$\frac{V_0^2{R}^2}{4r^2}=\frac{2GM}{r}-\frac{GM}{R}\Rightarrow \frac{\text{/}GM\text{/}{R}^2}{4\text{/}R\text{/}{r}^2}=\frac{2\text{/}GMR-\text{/}GMr}{R\text{/}r}$

$\frac{R}{4r}=\frac{2R-r}{R}\Rightarrow R^2=8Rr-4r^2$

$4r^2-8Rr+R^2=0\Rightarrow r=\frac{8R\pm \sqrt{64R^2-4\times 4R^2}}{2\times 4}$

$r=R+R\frac{\sqrt{3}}{2}\Rightarrow h=\frac{\sqrt{3}}{2}R$

and $V\ne \frac{V_0}{2}$

Path of the projectile will not be parabolic because the acceleration due to gravity is not constant throughout the motion.