## Conceptual Physics (12th Edition)

a. Find the time that the bullet is in the air, from the horizontal motion. $$d = vt = (400 \frac{m}{s})(t) = 200 m$$ $$t = 0.5 s$$ As far as vertical motion is concerned, the bullet started from rest. We need only find the height dropped in 0.5 s. $$y = \frac{1}{2}g t^{2} = 5 \frac{m}{s^{2}}(0.5 s)^{2} = 1.25 m$$ This is discussed on page 48. b. The bullet drops 1.25 m from its ideal straight-line path, so in this case the barrel should be aimed directly at a point 1.25 m above the bullseye.