Question

A ball is projected vertically upwards with an initial velocity of "u" goes to a maximum height "h" before coming to the ground. What is the value of h?

Answer 1

Consider upward direction as positive.

Initial velocity = u
at maximum height final velocity, v = 0
Acceleration, a = -g
Distance, s = h
Using the third equation of motion, we get
$v^2-u^2=2as$
$0-u^2=2(-g)h$
$u^2=2gh$
$h=u^2/2g$

Therefore the ball goes to a maximum height of $u^2/2g$.