Answer: 0.28 m
Deceleration, a= -60 g
where g is the acceleration due to gravity[tex]=9.8 m/s^2[/tex]
So, [tex]a= -60\times 9.8 m/s^2=-588 m/s^2[/tex]
Using the equation of motion, we need to find the initial velocity,
[tex]v-u=at[/tex]
final velocity, [tex]v=0[/tex]
time, [tex]t= 31 ms=31\times 10^{-3}s[/tex]
[tex]0-u=-588m/s^2\times 31\times 10^{-3} s=18.23 m/s [/tex]
now, using third equation of motion,
[tex]s=ut+\frac{1}{2}at^2[/tex]
we can find out the distance traveled by the person before coming to complete stop.
[tex]s=18.23m/s\times 31\times 10^{-3}s+\frac{1}{2}(-588 m/s^2) \times (31\times 10^{-3} s)^2=0.56 m-0.28 m=0.28 m[/tex]
