Question

Question #76814

Answer 1

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+at` (I)
`d=v_it+1/2at^2` (II)
`2ad=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 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.