# An object is thrown vertically from a height of 6 m at  1 m/s. How long will it take for the object to hit the ground?

Feb 10, 2016

#### Answer:

$t = 1 , 0087 s$
t=0,872 s

#### Explanation:

$\text{if object moves downward:}$
$h = {v}_{i} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$ ;;$6 = 1 \cdot t + \frac{1}{2} \cdot 9 , 81 \cdot {t}^{2}$
$6 = t + 4 , 905 \cdot {t}^{2}$
$4 , 905 \cdot {t}^{2} + t - 6 = 0$;$a {x}^{2} + b x + c = 0$
$\Delta = \sqrt{{b}^{2} - 4 \cdot a \cdot c}$ ;$\Delta = \sqrt{1 + 4 \cdot 4 , 905 \cdot 6}$
$\Delta = \sqrt{118 , 72}$
$\Delta = 10 , 896$
$\textcolor{red}{t = \frac{- b - \Delta}{2} \cdot a} \text{ time is not negative }$
$\textcolor{g r e e n}{t = \frac{- b + \Delta}{2} \cdot a}$
$t = \frac{- 1 + 10 , 896}{2 \cdot 4 , 905}$;;$t = \frac{9 , 896}{9 , 81}$
$t = 1 , 0087 s$
$\text{if object moves upward:}$
${t}_{u} = {v}_{i} / g = {t}_{u} = \frac{1}{9 , 81}$
$t = 0 , 1019 s \text{ elapsed time while object moves upward}$
${h}_{u} = {V}_{i}^{2} / \left(2 \cdot g\right) = \frac{1}{19 , 62} = 0 , 051 m \text{ height from starting point}$
$h = 6 + 0 , 051 = 6 , 051 m \text{ height to ground}$
$h = \frac{1}{2} \cdot g \cdot {t}^{2} \text{ freely falling}$
6,051=1/2*9,81*t^2 ; t=sqrt((2*6,051)/(9,81)) t=0,872 s#