Answer:
23.7402 m
21.582 m/s
310.521816 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m[/tex]
The drop distance is 23.7402 m
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s[/tex]
When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s
The final speed of the fall will be the initial velocity of stopping
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2[/tex]
The average stopping acceleration is 310.521816 m/s²