Question #76814

Jan 7, 2015

This is a complicated and easy question at the same time.
Easy because you can use the same equations you use for normal motion (say, under the influence of acceleration):
${v}_{f} = {v}_{i} + a t$ (I)
$d = {v}_{i} t + \frac{1}{2} a {t}^{2}$ (II)
$2 a d = {v}_{f}^{2} - {v}_{i}^{2}$ (III)

Difficult because you have a "fixed" acceleration, the acceleration of gravity, $g$.
This acceleration is always directed towards the centre of our planet so, when you analyze problems involving vertical motion you must be careful with signs.
What I do is to set a reference frame where "down" is negative and "up" is positive.
As a consequence the $a$ in equations (I), (II) and (III) has always the negative value of $- 9.81 \frac{m}{{s}^{2}}$.

If you throw a stone upward (from the origin at $y = 0$) with initial velocity ${v}_{i}$ (positive) gravity will reduce it ($g$ is downward, negative) until reaching ${v}_{f} = 0$ at the top (positive height $d$).
The stone will subsequently accelerate downwards under the action of $g$ acquiring negative velocity (downwards).

Hope it helped.