A projectile is fired straight up from the surface of the earth with an initial velocity vi ….

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}$

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Ans: (a)

$\displaystyle -\frac{G M m}{R} + \frac{1}{2} m v_i^2 = -\frac{G M m}{R+h}$

$\displaystyle \frac{1}{2} m v_i^2 = \frac{G M m}{R} -\frac{G M m}{R+h}$

$\displaystyle \frac{1}{2} m v_i^2 =G M m [ \frac{1}{R} -\frac{1}{R+h}]$

$\displaystyle \frac{1}{2} v_i^2 = G M [ \frac{1}{R} -\frac{1}{R+h}]$

$\displaystyle \frac{1}{2} v_i^2 = G M (\frac{h}{R(R+h)})$

$\displaystyle \frac{1}{2} v_i^2 = G M (\frac{h}{R h(\frac{R}{h} + 1)})$

$\displaystyle \frac{1}{2} v_i^2 = g R^2 (\frac{1}{R(\frac{R}{h} + 1)})$

$\displaystyle v_i^2 = 2 g R (\frac{1}{(1 + \frac{R}{h})})$

$\displaystyle v_i^2 = 2 g R (1 + \frac{R}{h})^{-1}$

$\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} $