a balloon is 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. after how much time and with what velocity does it reach the ground?
Answers
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.