Kinematics , what is the initial velocity

A football is thrown at 40 degrees to a guy that is 31.5 meters away. What must the initial velocity be?

Organize your work so i can understand how you did it. I want to know how to do it more than i want the answer.

R = (v^2 sin2A) / g

R = range = 31.5 m
v = initial velocity
A = angle of velocity = 40 degrees
g = gravity = 9.8 m/s^2

therefore,

v^2 = (Rg) / sin2A
v^2 = (31.5 * 9.8) / sin80
v^2 = 313.46
v = 17.7 m/s

You don't really need to show any work on this one, if the ball is being thrown then the initial velocity would be zero.
the variables you would want to find are acceleration, final velocity, and change in time.

The range equation:
R = V²sin(2Θ)/g
Solve it for V:

V = √[rg/sin(2Θ] = √[31.5*9.8/(sin80°)] = 17.7 m/s