Q: A projectile is fired straight up from the surface of the earth with an initial velocity vi . Neglecting air resistance, the ratio of radius of the earth and the maximum height up to which the projectile rises is
(a) $1 – \frac{v_i^2}{2 g R}$
(b) $ \frac{v_i^2}{2 g R}$
(c) $ \frac{v_i^2}{2 m g R}$
(d) $ \frac{v_i^2}{m g R}$
Click to See Answer :
$\displaystyle -\frac{G M m}{R} + \frac{1}{2} m v_i^2 = -\frac{G M m}{R+h}$
$\displaystyle \frac{1}{2} m v_i^2 = \frac{G M m}{R} -\frac{G M m}{R+h}$
$\displaystyle \frac{1}{2} m v_i^2 =G M m [ \frac{1}{R} -\frac{1}{R+h}]$
$\displaystyle \frac{1}{2} v_i^2 = G M [ \frac{1}{R} -\frac{1}{R+h}]$
$\displaystyle \frac{1}{2} v_i^2 = G M (\frac{h}{R(R+h)})$
$\displaystyle \frac{1}{2} v_i^2 = G M (\frac{h}{R h(\frac{R}{h} + 1)})$
$\displaystyle \frac{1}{2} v_i^2 = g R^2 (\frac{1}{R(\frac{R}{h} + 1)})$
$\displaystyle v_i^2 = 2 g R (\frac{1}{(1 + \frac{R}{h})})$
$\displaystyle v_i^2 = 2 g R (1 + \frac{R}{h})^{-1}$
$\displaystyle v_i^2 = 2 g R (1 – \frac{R}{h}) $
$\displaystyle 1 – \frac{R}{h} = \frac{v_i^2}{2 g R} $
$\displaystyle \frac{R}{h} = 1 – \frac{v_i^2}{2 g R} $