A projectile is launched at a certain speed at an angle of 23.8 degrees from the horizontal. Find the ratio o?

Position formula:
P(t) = 0.5 * a * t^2 + Vo * t + Po

For the vertical position, assuming the ground is level, then
  P(t) = P = 0
  Vo = Vvo = S [launch speed] * sin(23.8)
  a = -9.8 { acceleration due to gravity, negative is downward }

0 = 0.5 * (-9.8) * t^2 + Vo * t + 0
0 = t * (-4.9 * t + Vvo))
0 = t {the initial position}
-- OR --
Vvo / 4.9 = t

Ymax happens halfway through the flight at t/2 = Vvo / 9.8

Range uses the same position formula, but a = 0, Vo = Vho = S * cos(23.8), Po = 0. P(t), evaluated at t is the range.
P(t) = Vh0 * t

Putting it together:
0.5 * (-9.8) * (Vvo / 9.8)^2 + Vvo * (Vvo / 9.8)
-------------------------------------------------------------
Vho * (Vvo/4.9)

   -4.9 * Vvo^2 / 9.8^2 + Vvo^2 / 9.8
= ----------------------------------------------------------
   Vho * Vvo / 4.9

   (-1/2)(Vvo^2 / 9.8) + (Vvo^2 / 9.8)
= ----------------------------------------------------------
   Vho * Vvo / 4.9

   (-1/2 + 1)(Vvo^2 / 9.8)
= ----------------------------------------------------------
   Vho * Vvo / 4.9

   Vvo
= ----------------------------------------------------------
   4 * Vho

   S * sin(23.8)
= ----------------------------------------------------------
   4 * S * cos(23.8)

   sin(23.8)
= ----------------------------------------------------------
   4 * cos(23.8)