#### Answer

The ball is in the air for 3.2 seconds before it hits the ground.

#### Work Step by Step

$y = y_0+v_0~t+\frac{1}{2}at^2$
$0 = 2.0~m+(15~m/s)~t+\frac{1}{2}(-9.80~m/s^2)~t^2$
$(4.90~m/s^2)~t^2-(15~m/s)~t-2.0~m = 0$
We can then use the quadratic formula to find $t$:
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(-15)\pm \sqrt{(-15)^2-(4)(4.90)(-2.0)}}{(2)(4.90)}$
$t = 3.2~s, -0.13~s$
Since the negative value is unphysical, the solution is $t = 3.2~s$.
The ball is in the air for 3.2 seconds before it hits the ground.