# An object is thrown vertically from a height of 9 m at  2 m/s. How long will it take for the object to hit the ground?

##### 1 Answer

time $t = 1.574623134 \text{ }$seconds if thrown upward
time $t = 1.166459869 \text{ }$seconds if thrown downward

#### Explanation:

The formula for free falling bodies

$s = {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$
given $s = 9 \text{ }$meters
${v}_{0} = 2 \text{ }$meter/second

There are two solutions for the problem
(1) If the object is thrown vertically upward
$s = {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$
$- 9 = \left(2\right) \cdot t + \frac{1}{2} \left(- 9.8\right) \cdot {t}^{2}$

$4.9 {t}^{2} - 2 t - 9 = 0$

Using quadratic formula

$t = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(4.9\right) \left(- 9\right)}}{2 \left(4.9\right)}$

${t}_{1} = + 1.574623134 \text{ }$seconds
${t}_{2} = - 1.166459869 \text{ }$seconds

(2) If the object is thrown vertically downward
$s = {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$
$- 9 = \left(- 2\right) \cdot t + \frac{1}{2} \left(- 9.8\right) \cdot {t}^{2}$

$4.9 {t}^{2} + 2 t - 9 = 0$

Using quadratic formula

$t = \frac{- \left(+ 2\right) \pm \sqrt{{\left(+ 2\right)}^{2} - 4 \left(4.9\right) \left(- 9\right)}}{2 \left(4.9\right)}$

${t}_{1} = + 1.166459869 \text{ }$seconds
${t}_{2} = - 1.574623134 \text{ }$seconds

God bless....I hope the explanation is useful.