# An object is thrown vertically from a height of 12 m at  19 ms^-1. How long will it take for the object to hit the ground?

Jan 23, 2016

Assuming the object was thrown downward, the time taken will be $0.55$ $s$.

#### Explanation:

We know:

$u = 19$ $m {s}^{-} 1$
$d = 12$ $m$
$a = 9.8$ $m {s}^{-} 2$
t=? $s$

(we are choosing downward as the positive direction, since both the initial velocity and the acceleration are in that direction)

Use $d = u t + \frac{1}{2} a {t}^{2}$

This one is not so easy to rearrange to have $t$ as the subject, so let's just substitute in what we know:

$d = u t + \frac{1}{2} a {t}^{2} \to 12$
$19 \cdot t + \frac{1}{2} \cdot 9.8 \cdot {t}^{2}$

Rearrange the quadratic into standard form:

$4.9 {t}^{2} + 19 t - 12 = 0$

Solve the quadratic equation using the quadratic formula, or whichever other method you prefer. This yields two roots, but one makes more sense than the other:

$t = - 4.4$ $s$ or $t = 0.55$ $s$