Given data : A balloon is ascending at the rate of 9.8 m/s at a height of 39.2 m above the ground when a food packet is dropped from the balloon.

To find : After how much time and with what velocity does food packet reach the ground ?

Solution : Balloon is moving in an upward direction which means it is opposite to gravitational force, hence

⟹ Initial velocity of food packet, u = - 9.8 m/s

Here, we know that acceleration due to gravity, g = 9.8 m/s² so acceleration of food packet is, a = 9.8 m/s².

According to given,

⟹ Displacement of food packet, s = 39.2 m

Now, by kinematical equation

⟹ s = ut + ½ at²

⟹ 39.2 = - 9.8 t + ½ * 9.8 t²

⟹ 39.2 = - 9.8 t + 4.9 t²

⟹ 4.9 t² - 9.8 t - 39.2 = 0

⟹ 4.9 ( t² - 2t - 8 ) = 0

⟹ t² - 2t - 8 = 0 * 4.9

⟹ t² - 2t - 8 = 0

⟹ t² - 4t + 2t - 8 = 0

⟹ t (t - 4) + 2 (t - 8) = 0

⟹ (t + 2) (t - 4) = 0

⟹ t + 2 = 0 or t - 4 = 0

⟹ t = - 2 sec or t = 4 sec

Here, we know that time is never negative,

∴ t ≠ - 2 sec and t = 4 sec

Hence, food packet take 4 sec to reach the ground.

By using kinematical equation to find final velocity of food packet.

⟹ v = u + at

⟹ v - u = at

⟹ v - (- 9.8) = 9.8 * 4

⟹ v + 9.8 = 39.2

⟹ v = 39.2 - 9.8

⟹ v = 29.4 m/s

Hence, velocity of food packet to reach the ground is 29.4 m/s.