Question

A packet is dropped from a balloon which is going upwards with the velocity $12\, m$ $s^{-1}$, the velocity of the packet after 2 s will be

• A. $-12 \,m$ $s^{-1}$
• B. $12\, m$ $s^{-1}$
• C. $-7.6\, m$ $s^{-1}$
• D. $7.6\, m$ $s^{-1}$

Answer 1

Answer: (C)

$-7.6\, m$ $s^{-1}$

Answer 2

Given velocity of balloon u = $12 ms ^{-1}$ Times t=2 s when the packet is released from the balloon it aquires the velocity of balloon it means initial velocity of the packet = $12 ms^{-1}$ Hence, velocity after 2 s is given by, v = u + at = $12+(-9.8) \times 2$ =$12-19.6 = -7.6 ms^{-1}$ (minus sign in due to the motion is in opposite direction)