A marble is to be thrown horizontally from a height of $19.6\text{cm}$ above the ground so that it hits another marble on the ground $2\text{m}$ away. The velocity with which the marble should be thrown is

The correct option is B $10{\text{ms}}^{-1}$
Initial height of the marble from the ground $=19.6\text{cm}$
Using second equation of motion,
$h=u_{y}t+\frac{1}{2}g{t}^2\text{Since}{u}_y=0,t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times \frac{19.6}{100}}{9.8}}\therefore t=0.2s$
Horizontal distance covered by the marble $=R=2\text{m}$
$\therefore$ Horizontal velocity with which the marble is thrown
$v=\frac{R}{t}=\frac{2\text{m}}{0.2\text{s}}=10{\text{ms}}^{-1}$

Alternate solution:
Horizontal distance covered by a body falling from height $h$ with initial horizontal velocity $u$
$X=u\sqrt{\frac{2h}{g}}$
i.e. $2=u\sqrt{\frac{2\times 19.6}{9.8\times 100}}$
$\Rightarrow u=10{\text{ms}}^{-1}$