(a) The horizontal distance covered by the bomb,
$BC = \upsilon_{H} \times \sqrt{\frac{2h}{g}} = 150\sqrt{\frac{2 \times 80}{10}} = 600\ m$
$\therefore$ The distance of target from dropping point of bomb,
$AC = \sqrt{AB^{2} + BC^{2}\ } = \sqrt{(80)^{2} + (600)^{2}} = 605.3\ m$