# An object is thrown vertically at a height of 14 m at 1 m/s. How long will it take for the object to hit the ground?

Apr 12, 2016

$t = 1 , 59 \text{ s}$
$t = 1 , 69 \text{ s}$

#### Explanation:

$\text{ if object is thrown downward:}$

${v}_{i} = 1 \frac{m}{s}$
$y = 14 m$
$g = 9 , 81 \frac{m}{s} ^ 2$

$y = {v}_{i} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$
$14 = 1 \cdot t + \frac{1}{2} \cdot 9 , 81 \cdot {t}^{2}$

$4 , 905 {t}^{2} + t - 14 = 0$

$\Delta = \sqrt{{1}^{2} + 4 \cdot 4 , 905 \cdot 14}$

$\Delta = \sqrt{1 + 274 , 68}$

$\Delta = \sqrt{275 , 68}$
$\Delta = 16 , 60$
$t = \frac{- 1 + 16 , 60}{2 \cdot 4 , 905}$

$t = \frac{15 , 60}{9 , 81}$
$t = 1 , 59 \text{ s}$
$\text{if object is thrown upward :}$
${t}_{u} = {v}_{i} / g \text{ "t_u=1/(9,81)" "t_u=0,10" s}$
$\text{elapsed time to reach peak point}$
$h = {v}_{i}^{2} / \left(2 \cdot g\right)$

$h = \frac{1}{2 \cdot 9 , 81} \text{ "h=0,05" m}$
${h}_{t} = 14 + 0 , 05 = 14 , 05 \text{ meters total height}$

$14 , 05 = \frac{1}{2} \cdot g \cdot {t}^{2} \text{ } 28 , 1 = 9 , 81 \cdot {t}^{2}$

${t}^{2} = \frac{28 , 1}{9 , 81}$

${t}^{2} = 2 , 86$

$t = 1 , 69 \text{ s}$