A ball after freely falling from a height of $4.9\text{m}$ strikes a fixed horizontal plane. If the coefficient of restitution is $\frac{3}{4}$, the ball will strike second time with the plane after

The correct option is C $1.5\text{s}$
According to question,

Velocity of ball just before striking plane, $v$
using,
$v^2-u^2=2as$
$\Rightarrow v=\sqrt{2\times 9.8\times 4.9}=9.8\text{m/s}$ ...(1)
Since horizontal plane remains fixed, initial and final velocity of the plane remains same.
Rebound velocity, $v^{\prime }=ev$
$=\frac{3}{4}\times v$ [$e=3/4$, given]
$=\frac{3}{4}\times 9.8$ [from (1)]
$=7.35\text{m/s}$..... (2)
Let us suppose, after ${}^{\prime }{t}^{\prime }$ time ball again strikes the plane
As ball reaches to same position, its displacement is zero.
Using, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 0=7.35t-\frac{1}{2}\times 9.8\times t^2$ [from (2), $u=v^{\prime }$]
$\Rightarrow t=\frac{7.35\times 2}{9.8}$
$\Rightarrow t=\frac{3}{2}=1.5\text{s}$
After $1.5\text{s}$ ball will strike the plane for the second time.