Answer:
The position is  [tex]P = 47.4 \ m[/tex] relative to the base of the ocean
Explanation:
From the question we are told that
  The angle made by the incline with the horizontal is [tex]\theta = 24.0 ^o[/tex]
  The constant acceleration is [tex]a = 3.82 \ m/s^2[/tex]
  The distance covered is [tex]d = 60.0 \ m[/tex]
  The height of the cliff is [tex]h = 50 .0 \ m[/tex]
The velocity of the car is mathematically represented as
   [tex]v^2 = u^2 + 2ad[/tex]
The initial velocity of the car is  u= 0
So
   [tex]v^2 = 2ad[/tex]
substituting values
   [tex]v^2 = 2 * 3.82 * 60[/tex]
  [tex]v = 21.4 \ m/s[/tex]
The vertical component of this velocity is
  [tex]v_v = -v * sin(\theta )[/tex]
substituting values
  [tex]v_v = -21.4 * sin(24.0)[/tex]
  [tex]v_v = -8.7 \ m/s[/tex]
The negative sign is because is moving in the negative direction of the y-axis
The horizontal  component of this velocity is
   [tex]v_h = v * cos (\theta)[/tex]
  [tex]v_h = 21.4 * cos (24.0)[/tex]
  [tex]v_h = 19.5 \ m/s[/tex]
Now according to equation of motion we have
   [tex]h = v_v*t - \frac{1}{2} * g t^2[/tex]
substituting  values
  [tex]50 = -8.7 t - \frac{1}{2} * 9.8 t^2[/tex]
  [tex]4.9t^2 +8.7t -50 = 0[/tex]
using quadratic equation we have that
 [tex]t_1 = 2.42\ s \ and\ t_2 = -4.20\ s[/tex]
given that time cannot be negative
   [tex]t = 2.42 \ s[/tex]
The  car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as
      [tex]P = v_h * t[/tex]
substituting values
     [tex]P = 19.5 * 2.43[/tex]
     [tex]P = 47.4 \ m[/tex]
