in a movie, a monster climbs to the top of a building 40 meters above the group and hurls a boulder downward with a speed of 28 m/s at an angle of 48 degrees below the horizontal. how far from the building does the boulder land?

For the final time: - h = v_0, y * t - g*t^2/2 we have a quadratic equation g*t^2/2 - v_0, y * t + h = 0 whose solutions are t = 2*(v_0, y +/- sqrt{(v_0, y)^2 + 2*g*h})/g = 2*(-20.8 +/- sqrt{20.8^2 + 2*9.81*40}) = 1.47 s.

In the last formula we took only plus sign since the solution with minus sign is negative and does not make sense. Finally, x = v_0, x * t = 18.7 * 1.47 = 27.5 m.

Answer: The boulder will land 27.5 m away from the building.