Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$y = y_0+v_0~t+\frac{1}{2}at^2$ $0 = 2.0~m+(15~m/s)~t+\frac{1}{2}(-9.80~m/s^2)~t^2$ $(4.90~m/s^2)~t^2-(15~m/s)~t-2.0~m = 0$ 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.