Question

A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.7 m/s . When the balloon is 21 m above the ground, she accidentally releases the bag. Part A What is the speed of the bag just before it reaches the ground

Answer 1

Answer:

The speed of the bag just before it reaches the ground is 21.71 m/s.

Explanation:

The speed of the bag can be found using the following equation:

$v_{f}^{2} = v_{0}^{2} - 2gy$

Where:

$v_{f}$ is the final speed =?

$v_{0}$ is the initial speed = 7.7 m/s

g: is the gravity = 9.81 m/s²

y: is the height = 21 m

$v_{f}^{2} = v_{0}^{2} - 2gy = (7.7 m/s)^{2} - 2*9.81 m/s^{2}*(-21 m) = 471.31 m^{2}/s^{2}$

$v_{f} = 21.71 m/s$

Therefore, the speed of the bag just before it reaches the ground is 21.71 m/s.

I hope it helps you!