Question

A projectile is fired vertically upwards from the surface of the Earth with a velocity $k\upsilon _e$ , where $\upsilon_e$ is the escape velocity and $k < 1$ . If $R$ is the radius of the Earth, the maximum height to which it will rise measured from the centre of the Earth will be

• A. $\frac{1 - k^2}{R}$
• B. $\frac{R}{1 - k^2}$
• C. $R (1 - k^2)$
• D. $\frac{R}{1 + k^2}$

Answer 1

Answer: (B)

$\frac{R}{1 - k^2}$

Answer 2

From law of conservation of energy,
$\frac{1}{2} m \upsilon^2 = \frac{mgh}{1 + \frac{h}{R}}$
$\because \, \upsilon = k \upsilon_2 = k \sqrt{2gR}$
$\therefore \, \frac{1}{2} m k^2 . 2g R = \frac{mg(r- R)}{1 + \frac{r -R}{R}}$
$k^2 R \left[ 1 + \frac{r - R}{R} \right] = r - R$
$k^2 r = r - R \, \Rightarrow r = \frac{R}{1 - k^2}$