Three particles are projected vertically upward from a point on the surface of the earth with velocities $\sqrt{(2 g R / 3)}, \sqrt{(g R)}, \sqrt{(4 g R / 3)}$ respectively where $R$ is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The maximum heights attained are respectively $h_{1}, h_{2,} h_{3}$

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$h_{1} : h_{2} = 2 : 3$

$h_{2} : h_{3} = 3 : 4$

$h_{1} : h_{3} = 1 : 4$

$h_{2} = R$

answer is C, D.

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Detailed Solution

If $V = K V_{e} (K < 1)$ , the maximum height
attained by the body $h = \frac{{RK}^{2}}{1 - K^{2}}$

$V = \sqrt{\frac{2 gR}{3}} = \frac{1}{\sqrt{3}} V_{e} \Rightarrow K = \frac{1}{\sqrt{3}}$

$∴ h_{1} = \frac{R (\frac{1}{3})}{1 - \frac{1}{3}} = R / 2$

$V = \sqrt{gR} = \frac{1}{\sqrt{2}} \sqrt{2 gR} \Rightarrow K = \frac{1}{\sqrt{2}}$

$∴ h_{2} = \frac{R (\frac{1}{2})}{1 - \frac{1}{2}} = R$

$V = \sqrt{\frac{4 gR}{3}} = \sqrt{\frac{2}{3}} \sqrt{2 gR} \Rightarrow K = \sqrt{2 / 3}$

$∴ h_{3} = \frac{R (\frac{2}{3})}{1 - \frac{2}{3}} = 2 R$ $∴ \frac{h_{1}}{h_{3}} = \frac{1}{4} & h_{2} = R$

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