Answer:
[tex]a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Explanation:
According to Newton's Third Equation of motion
[tex]V_f^2-V_i^2=2as[/tex]
Where:
[tex]V_f[/tex] is the final velocity
[tex]V_i[/tex] is the initial velocity
a is the acceleration
s is the distance
In our case:
[tex]V_f=V_{escape}, V_i=0,s=4 m[/tex]
So Equation will become:
[tex]V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]