Q. A stone is thrown vertically upward with an initial velocity *v*_{0}. The distance travelled in time $\displaystyle \frac{1.5v_0}{g}$ is

(a) $ \displaystyle \frac{{v_0}^2}{2 g}$

(b) $\displaystyle \frac{3{v_0}^2}{8 g}$

(c) $ \displaystyle \frac{5{v_0}^2}{8 g}$

(d) None of these

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$\displaystyle h = \frac{v_0^2}{2g}$

Time taken by stone to attained maximum height is

$ \displaystyle t_1 = \frac{v_0}{g}$

Remaining time $\displaystyle t_2 = \frac{1.5v_0}{g} – \frac{v_0}{g} $

$ \displaystyle t_2 = \frac{v_0}{2g}$

Distance covered in time t_{2} during downward journey

$\displaystyle h’ = \frac{1}{2}g t_2^2$

$ \displaystyle h’ = \frac{1}{2}g (\frac{v_0}{2g})^2$

$ \displaystyle h’ = \frac{v_0^2}{8g}$

Total distance covered = h + h’

$ \displaystyle = \frac{v_0^2}{2g} + \frac{v_0^2}{8g}$

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