Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
[tex]v^2=u^2+2as[/tex]
[tex]v^2=(7.3)^2+2\times 9.8\times 24[/tex]
[tex]v=\sqrt{523.69}[/tex]
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
