How do I do this 1D Kinematics question?

Question

A ball is thrown upwards with an inital speed v. Show that the maximum height reached by the ball is v^2/2g and that the time taken for it to return to the ground is 2v/g.

Answer 1

let initial speed be u

speed at maximum height(v) is 0

a^= -g

now v^2 = u^2 + 2as

0 = u^2 -2gs

2gs = u^2

s = u^2 /2g

s = ut + 0.5at^2

s = 0 as the displacement at the bottom is 0 due to being in starting position

0 =ut - 0.5gt^2

0.5gt^2 = ut

0.5gt^2 - ut = 0

t(0.5gt -u)

ball is in original position when t = 0,

and when 0.5gt -u = 0

0.5gt = u

gt = 2u

t = 2u/g

sorry but i am used to u being initial speed!