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

Updated On: Jul 5, 2022
  • $-12 \,m$ $s^{-1}$
  • $12\, m$ $s^{-1}$
  • $-7.6\, m$ $s^{-1}$
  • $7.6\, m$ $s^{-1}$
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The Correct Option is C

Solution and Explanation

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)
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