# An object is thrown vertically from a height of 2 m at 2 m/s. How long will it take for the object to hit the ground?

Apr 12, 2017

0.56518s

#### Explanation:

For a distance, $d$, an initial velocity, ${v}_{0}$, an acceleration of $a$, and a time of $t$
$\textcolor{w h i t e}{\text{XXX}} d = {v}_{0} t + a {t}^{2}$

In this case
color(white)("XXX")d=-2 " m" (2 meters below the height from with it was thrown)If it was dropped and not thrown up there would be no negative sign
color(white)("XXX")v_0= 2 " m"/"sec"
$\textcolor{w h i t e}{{\text{XXX")a=-9.8 " m"/(sec}}^{2}}$ (standard gravity assumed)

So we have
color(white)("XXX")-2.0" m" = 2.0t " m" + (-9.8t^2)" m"

$\textcolor{w h i t e}{\text{XXX}} 98 {t}^{2} - 20 t - 20 = 0$

$\textcolor{w h i t e}{\text{XXX}} t = \frac{20 \pm \sqrt{{20}^{2} - 4 \left(98\right) \left(- 20\right)}}{2 \left(98\right)}$
$\textcolor{w h i t e}{\text{XXX}} t = 0.56518 s$