Q. A stone is thrown vertically upward with an initial velocity v0. 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
Click to See Answer :
$\displaystyle h = \frac{v_0^2}{2g}$
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 t2 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} $