A rock is thrown at a window that is located 16.0 m above the ground. The rock is thrown from the ground at an angle of 40.0° above horizontal with an initial speed of 30.0 m/s and experiences no appreciable air resistance. If the rock strikes the window on its upward trajectory, from what horizontal distance from the window was it released?

x = 27.3 m

Explanation:

This is a projectile launching exercise, let's start by looking for the time it takes for the rock to reach the height of the window.

Let's use trigonometry to find the velocities of the rock

           sin 40 = / v

           cos 40 = v₀ₓ / v

           v_{oy}= v sin 40

           v₀ₓ = v cos 40

           v_{oy} = 30 sin 40 = 19.28 m / s

           v₀ₓ = v cos 40

           v₀ₓ = 30 cos 40 = 22.98 m / s

we look for the time

        = v_{oy}^2 - 2 g y

         v_{y}^2 = 19.28 2 - 2 9.8 16 = 371.71 - 313.6 = 58.118

        v_{y} = 7.623 m / s

we calculate the time

          v_{y} = v_{oy} - gt

          t = (v_{oy} - v_{y}) / g

          t = (19.28 -7.623) / 9.8

          t = 1,189 s

since the time is the same for both movements let's use this time to find the horizontal distance

           x = v₀ₓ t

           x = 22.98 1,189

           x = 27.3 m

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