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(C) The maximum height reached by a body thrown with initial velocity $u$ is given by $H = \frac{u^2}{2g}$,which implies $u^2 = 2gH$.Using the equatio

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$3: 2$

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(C) The maximum height reached by a body thrown with initial velocity $u$ is given by $H = \frac{u^2}{2g}$,which implies $u^2 = 2gH$.Using the equation of motion $v^2 = u^2 - 2gh$,we calculate the velocity at any height $h$.For height $h_1 = \frac{3H}{4}$,the velocity $v_1$ is given by $v_1^2 = u^2 - 2g(\frac{3H}{4}) = 2gH - \frac{3gH}{2} = \frac{gH}{2}$.For height $h_2 = \frac{8H}{9}$,the velocity $v_2$ is given by $v_2^2 = u^2 - 2g(\frac{8H}{9}) = 2gH - \frac{16gH}{9} = \frac{2gH}{9}$.The ratio of the velocities is $\frac{v_1}{v_2} = \sqrt{\frac{v_1^2}{v_2^2}} = \sqrt{\frac{gH/2}{2gH/9}} = \sqrt{\frac{1}{2} 	imes \frac{9}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.

$A$ body thrown vertically upwards from the ground reaches a maximum height $H$. The ratio of the velocities of the body at heights $\frac{3H}{4}$ and $\frac{8H}{9}$ from the ground is

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