Question #76814

1 Answer

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.