The speed that the person would hit the ground is equal to [tex]7.39\frac{m}{s}[/tex]
Why?
From the statement we know that the person steps off accidentally, so, we can safely assume that there is no initial speed. We can calculate the final speed (the speed at which the person would hit the ground)using the following formula:
[tex]v^{2}=v_{o}^{2} +2g*d[/tex]
Where,
v is the final speed
[tex]v_{o}[/tex] is the initial speed (considered zero since the person is falling)
g is the acceleration due to gravity (considered positive since the person is falling)
d is the height of the stepladder
In order to use the formula, we need to convert from ft. to meters, so, converting we have:
[tex]1m=3.28ft[/tex]
[tex]9f*\frac{1m}{3.28ft}=2.74m[/tex]
Now, substituting the given information and calculating we have:
[tex]v^{2}=v_{o}^{2} +2g*d\\\\v^{2}=0^{2} +2*(9.81\frac{m}{s^{2}})*(2.7m)\\\\v^{2}=2*(9.81\frac{m}{s^{2}})*(2.7m)=19.92\frac{m}{s^{2}}*2.74m=54.58\frac{m^{2} }{s^{2}} \\\\v=\sqrt{54.58\frac{m^{2} }{s^{2}}}=7.38\frac{m}{s}[/tex]
Hence, the speed at which the person would hit the ground is equal to [tex]7.39\frac{m}{s}[/tex]
Have a nice day!
