# An object is thrown vertically from a height of 15 m at  19 m/s. How long will it take for the object to hit the ground?

Feb 10, 2016

$4.55 s$

#### Explanation:

The vertical distance the object travels is given by the equation
$y = {v}_{y 0} t - \frac{1}{2} g \cdot {t}^{2}$ where ${v}_{y 0}$ is the vertical component of the initial velocity, $g$ is the acceleration due to gravity and $t$ is the time taken to reach the highest point (where the velocity is zero).
The time can be found from the equation ${v}_{y f} = {v}_{y o} - g \cdot t$ by using the fact that ${v}_{y f} = 0$

Thus $0 = 19 - 9.8 \cdot t$
$t = \frac{19}{9.8} = 1.94 s$

The height the object reaches is therefore $19 t - 9.8 \cdot {t}^{2} / 2$
$= 18.4 m$

We now have to work out how long the object will take to fall $18.4 + 15 = 33.4 m$

$33.4 = 0 \cdot t + \frac{1}{2} g \cdot {t}^{2}$
$t = \sqrt{33.4 \cdot \frac{2}{9.8}} = 2.61 s$

The total time is $1.94 + 2.61 = 4.55 s$