Answer
The object falls in a time of $~~3.41~s$
Work Step by Step
We can write an expression for $h$ in terms of $t$:
$h = \frac{1}{2}at^2$
$h = 4.9~t^2$
Let $v$ be the velocity after the object falls a distance of $0.5~h$
We can find an expression for $v$:
$v^2 = v_0+2ay$
$v^2 = 0+(2)(9.8)(0.5~h)$
$v = \sqrt{9.8~h}$
We can write an equation for the final $0.5~h$:
$0.5~h = v~t+\frac{1}{2}at^2$
$0.5~h = (\sqrt{9.8~h})(1.00)+(4.9)(1.00)^2$
$0.5~h = \sqrt{9.8~h}+4.9$
$(0.5)~(4.9~t^2) = \sqrt{(9.8)~(4.9~t^2)}+4.9$
$2.45~t^2 = 6.93~t+4.9$
$2.45~t^2 - 6.93~t-4.9 = 0$
We can use the quadratic formula:
$t = \frac{-(-6.93)\pm \sqrt{(-6.93)^2-(4)(2.45)(-4.9)}}{2(2.45)}$
$t = \frac{6.93\pm \sqrt{96.0449}}{4.9}$
$t = -0.586~s, 3.41~s$
Since time is positive, the solution is $t = 3.41~s$
The object falls in a time of $~~3.41~s$
