Time it takes bag to reach the ground

Question

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.05 m/s. Knowing that at the time of the release the balloon was 56.3 m above the ground, determine the time, τ, it takes the bag to reach the ground from the moment of its release.

Answer 1

x_i = 56.3 m

x_f = 0 m

v_i = 2.05 m/s

a = g = -9.8 m/s²

 

x_f = x_i + v_it + (1/2)at²

⇒ (0m) = (56.3m) + (2.05m/s)t + (1/2)(-9.8 m/s²)t²

⇒ 0 = 56.3m + (2.05m/s)t - (4.9m/s²)t²

 

Use quadratic equation to solve:

(-b ± (b² - 4ac)^0.5) / 2a

a = -4.9

b = 2.05

c = 56.3

⇒ (-(2.05) ± ((2.05)² - 4(-4.9)(56.3))0.5) / 2(-4.9)

⇒ (-2.05 ± 33.281) / (-9.8)

⇒ 0.20918 ± 3.3960

 

Since it doesn't make sense for time to be negative here, then the answer is:

0.20918 + 3.3960 = 3.6052s ≈ 3.60s

 

Answer 2

A_G = 9.81m/s²

Δx = -56.3m

V₀ = 2.05m/s

t = ?

 

 

Δx = V₀t + ¹/₂A_Gt²

 

-56.3 = 2.05t + ¹/₂(-9.81)t²

 

-56.3 = 2.05t + -4.905t²

 

+56.3                  +56.3

 

-4.905t² + 2.05t + 56.3 = 0

 

 

Use quadratic equation to solve for t:

 

(-2.05 - √2.05² - 4(-4.905)(56.3) ) / 2(-4.905)

 

(-2.05 - √4.2025 + 1104.606 ) / -9.81

 

(-2.05 - √1108.8085 ) / -9.81

 

(-2.05 - 33.2988) / -9.81

 

-35.3 / -9.81 = 3.60s

 

So it will take 3.60s to fall