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:
[tex] v_{f}^{2} = v_{0}^{2} - 2gy [/tex]
Where:
[tex]v_{f}[/tex] is the final speed =?
[tex]v_{0}[/tex] is the initial speed = 7.7 m/s
g: is the gravity = 9.81 m/s²
y: is the height = 21 m
[tex]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}[/tex]
[tex]v_{f} = 21.71 m/s[/tex]
Therefore, the speed of the bag just before it reaches the ground is 21.71 m/s.
I hope it helps you!
