Projectile Motion Problem

A golf ball is driven at a speed of 75.0m/s at an angle of 35.0 degrees as measured from the ground. The gold ball lands 15.0m below the elevation the ball started from. Calculate:

The vertical and horizontal component of the ball's velocity just before it hits the ground.
The velocity with which the ball hits the ground?
The Range of the ball?

I just need help getting started with the question.. how would I set it up to solve it? The only part confusing me is that the ball lands 15.0 m below the elevation it starts from :/ thanks in advance!

let time be t
vertical motion
s = ut + 0.5 at^2
- 15 = 75 sin 35 t - 0.5 * 9.81 * t^2
4.905 t^2 - 43.02 t - 15 =0
t = { 43.02 + sq root[2145] }/ 9.81 = 9.11s
t1 = { 43.02 - sq root[2145] }/ 9.81 not valid

The Range of the ball
= 75 * cos 35 * 9.11
m
answer
horizontal component of the ball's velocity just before it hits the ground
= 75 cos 35 m/s
answer
v= u +at
v = 75 sin 35 - 9.81 * 9.11 m/s
answer
vertical component of the ball's velocity just before it hits the ground

velocity with which the ball hits the ground
= square root [ vertical^2 + horizontal^2 component of the ball's velocity just before it hits the ground]
m/s
answer