Q. At the height $80\ m$, an aeroplane is moving with $150\ m/s$. A bomb is dropped from it so as to hit a target. At what distance from the target should be bomb be dropped $($Given$\ g = 10\ m/s^{2})$

A) $605.3\ m$
B) $600\ m$
C) $80\ m$
D) $230\ m$
Explanation

(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,
$AC = \sqrt{AB^{2} + BC^{2}\ } = \sqrt{(80)^{2} + (600)^{2}} = 605.3\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.