You are standing on a bridge that is 12 meters above the water. If you throw a rock with a speed of 12 m/s at an angle of 20 degrees above the horizontal, how far will it travel horizontally when it hits the water?

x = 22.99 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

          sin 20 = / v₀

          cos 20 = v₀ₓ / v₀

           v_{oy} = v₀ sin 20

           v₀ₓ = v₀ cos 20

            v_{oy} = 12 sin 20

            v₀ₓ = 12 cos 20

            v_{oy} = 4.1 m / s

             v₀ₓ = 11.28 m / s

let's look at the time it takes to reach the ground

            y = y₀ + v_{oy} t - ½ g t²

where i is the height of the bridge 12m

            0 = 12 + 4.1 t - ½ 9.8 t²

           t² - 0.837 t - 2.45 = 0

let's solve the quadratic equation

          t = [0.837 ±√ (0.837² +4 2.45)] / 2

          t = [0.837 ± 3.24] / 2

          t₁ = -1.2 s

          t₂ = 2.0385 s

as the time has to be positive the correct result is t = 2.04 s

we use this time to calculate the horizontal distance traveled

         x = v₀ₓ t

         x = 11.28 2.0385

         x = 22.99 m