#### Answer

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.