# An object is thrown vertically from a height of 9 m at 4 m/s. How long will it take for the object to hit the ground?

Jan 14, 2016

$t \approx 0.9 s$

#### Explanation:

Assuming no air resistance and using the equation
$s = u t + \frac{1}{2} g {t}^{2}$
Substituting given values and of $g = 9.8 \frac{m}{s} ^ 2$
$9 = 4 t + \frac{1}{2} 9.8 {t}^{2}$
Rearranging
$\frac{1}{2} 9.8 {t}^{2} + 4 t - 9 = 0$

$\implies$ $4.9 {t}^{2} + 4 t - 9 = 0$

Solving the quadratic equation using the formula
$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

and ignoring the $- v e$ value of t

$t = \frac{- 4 + 13.28}{9.8}$