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(C) The time taken to reach the maximum height is $t_0 = \frac{v_0}{g}$.Since the given time $t = \frac{1.5 v_0}{g}$ is greater than $t_0$,the stone r

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$\frac{5 v_0^2}{8 g}$

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(C) The time taken to reach the maximum height is $t_0 = \frac{v_0}{g}$.Since the given time $t = \frac{1.5 v_0}{g}$ is greater than $t_0$,the stone reaches the maximum height and then starts falling back.The distance travelled is the sum of the distance to the maximum height and the distance fallen from the maximum height.Distance to maximum height $h_{max} = \frac{v_0^2}{2g}$.Time spent falling back is $t' = t - t_0 = \frac{1.5 v_0}{g} - \frac{v_0}{g} = \frac{0.5 v_0}{g} = \frac{v_0}{2g}$.Distance fallen in time $t'$ is $d' = \frac{1}{2} g (t')^2 = \frac{1}{2} g \left(\frac{v_0}{2g}\right)^2 = \frac{v_0^2}{8g}$.Total distance = $h_{max} + d' = \frac{v_0^2}{2g} + \frac{v_0^2}{8g} = \frac{4v_0^2 + v_0^2}{8g} = \frac{5 v_0^2}{8g}$.

$A$ stone is thrown vertically upward with an initial velocity $v_0$. The distance travelled in time $t = \frac{1.5 v_0}{g}$ is

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