Verified answer

Answer:

Option (C) -7.6 m/s is correct.

Explanation:

Given that the packet was also moving up along with the balloon. Hence the velocity of both the packet and balloon must be the same.

Hence, u = 12 m/s and t = 2 sec

Since, the balloon must fly in the atmosphere, so it must experience a constant gravitational force towards the earth. After the packet is dropped, the only force acting upon the packet is gravity (neglecting air friction).

Hence, net acceleration of the packet is "g".

Also, the acceleration is towards downwards, so, we will take it with a negative sign i.e. a = g = –9.8m/s²

★ Formula used: v = u + at

★ Solution:

Now, using the first equation of motion

→ v = u + at

→ v = 12 + (–9.8) × 2

→ v = 12 – 19.6

→ v = –7.6 m/s

Hence, the magnitude of the final velocity of the particle is 7.6 m/s.

It is important to note that negative sign shows that the velocity is in the direction of "g" or opposite to the direction of movement of the balloon i.e. it’s direction is downwards.