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2.Motion in Straight Line
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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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