## Conceptual Physics (12th Edition)

a. Use the time of 30 seconds. As far as vertical motion is concerned, the engine started from rest. We need only find the height from which it takes 30 s to hit the ground. $$y = \frac{1}{2}g t^{2} = 5 \frac{m}{s^{2}}(30 s)^{2} = 4500 m$$ This is discussed on page 48. b. The horizontal velocity of the engine does not change. Find the horizontal distance it moved in the 30-s drop time. $$d = vt = (280 \frac{m}{s})(30 s) = 8400 m$$ c. The plane would have traveled the same horizontal distance, so in the absence of air resistance, the fallen engine lands directly below the airplane.