Gujarati
Hindi
2.Motion in Straight Line
hard

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

A

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

B

$\frac{3 v_0^2}{8 g}$

C

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

D

None of these

Solution

(c)

Velocity of particle will become zero in time $t_0=\frac{v_0}{g}$

The given time $t=\frac{1.5 v_0}{g}$ is greater than the time $t_0=\frac{v_0}{g}$

Hence, distance $ > \mid$ displacement $\mid$

Distance $=\left|S_{0-t_0}\right|+\left|S_{t-t_0}\right|$

$=\frac{v_0^2}{2 g}+\frac{1}{2} g\left(t-t_0\right)^2$

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

Standard 11
Physics

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