Q. An aeroplane moving horizontally with a speed of $720\ km/h$ drops a food pocket, while flying at a height of $396.9\ m$. The time taken by a food pocket to reach the ground and its horizontal range is $(Take\ g = 9.8\ m/\sec^{2}\ )$
A) $3\sec{\ and\ 2000\ m}$
B) $5\sec{\ and\ 500\ m}$
C) $8\sec{\ and\ 1500\ m}$
D) $9\sec{\ and\ 1800\ m}$
Explanation
(d) $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 396.9}{9.8}} \simeq 9\sec\ $ and $u = 720km/hr = 200\ m/s$
$\therefore R = u \times t = 200 \times 9 = 1800\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.