A ball is thrown from the brink of a cliff 17 m high with an initial velocity of v0 = 8 m/s at an?

A very easy question of "Projectiles".
First i'll introduce the letters.
The height of ball goes from the cliff: h
The height of the ball goes from base of the cliff: H ( H= h+ 17m )
The horizontal distance of the ball from the cliff : S
Total time ball takes to complete projectile : t
Total time ball takes to complete from end of the projectile to base of the cliff : t'
Total time ball takes to base of of the cliff : T ( T= t+ t')
The velocity of ball when it ends the movement of projectile : v

I took gravitational force as 10m/s
So this is how it solve.

To the 1st half movement of projectile > downwards,
v2 = u2 + 2as
0 = (8 sin 32 )2 - 2 x 10 x h
20h = (8 x 0.5299)2
20h = (4.2392)2
20h = 17.97m
h = 0.89 m

to the 2nd half movement of projectile > downwards,
S = ut + 1/2 a(t/2)2
h = 1/2 x 10 x (t/2)2
(t/2)2 = ( 2x 0.89 ) / 10
(t/2)2 = 0.178
t/2 = 0.42
so the total time ball takes for projectile is, t = 0.84seconds

to the 1st half of projectile > rightwards,
S = ut + 1/at2
S/2 = 8cos32t'
S/2 = 8x 0.8480 x 0.42
S = 5.69m

to the 2nd half of the projectile > downwards,
v2= u2 + 2aS
v2 = 0+ 2 x 10 x 0.89
v2 = 17.8
v = 4.21 m/s

to the gravitational movement of the ball from end of the projectile to the base of the cliff > downwards,
S = ut' + 1/2 a t'2
17 = 4.2t' + 1/2 x 10 x t'2
Find t'

so, for the Total time, T= t + t'
for the Total height ball goes, H = h + 17
the vertical distance ball goes, S .

:) :)