Kinematics Question 199
Question: An aeroplane is moving with a velocity $ u $ . It drops a packet from a height $ h $ . The time $ t $ taken by the packet in reaching the ground will be
Options:
A) $ \sqrt{( \frac{2g}{h} )} $
B) $ \sqrt{( \frac{2u}{g} )} $
C) $ \sqrt{( \frac{h}{2g} )} $
D) $ \sqrt{( \frac{2h}{g} )} $
Show Answer
Answer:
Correct Answer: D
Solution:
The initial velocity of aeroplane is horizontal, then the vertical component of velocity of packet will be zero. So $ t=\sqrt{\frac{2h}{g}} $
BETA
