Answer
total time during which ball remain in air $=t=3.28s $
Work Step by Step
When the ball will come back to its intial position that is 2m above from ground its velocity will be same but in opposite direction.
Net displacement will be zero.
let $t_1$ be the total time taken by the ball to reach to its intial height.
We have
$\vec s_1=0$
$\vec u=15m/s$
$\vec v=-15m/s$
$\vec a=-g =-9.8m/s^2$
We know that
$\vec s_1 = \vec ut_1 +\frac{1}{2}\vec a (t_1)^2$s
$0= 15t_1 +\frac{1}{2}(-9.8)(t_1)^2$s
$30t_1 =9.8( t_1)^2$s
$t_1 =\frac{30}{9.8} $s
$t_1 =3.06$s
let $t_2 $ be the time taken from 2m height to ground.
here $\vec s_2 =2m \\ \vec a= 9.8 m/s^2 \\ \vec u= 15 m/s$
$\therefore \vec s_2 = \vec ut_2 + \frac{1}{2}\vec a(t_2)^2$
$ 2 = 15 t _2+ 4.9 (t_2)^2$
$t_2 = 0.22$s
thus total time during which ball remain in air $=t=t_1+t_2 =3.06s + 0.22s =3.28s $
