Given,
The mass of the gymnast, m=40 kg
The initial speed with which the gymnast enters the foam ball pit, u=7 m/s
The average resistive force applied by the foam balls, F=-1000 N
The negative sign indicates that the force is resistive in nature.
From Newton's second law,
[tex]F=ma[/tex]
Where a is the acceleration with which the gymnast slows down after entering the pit.
On substituting the known values,
[tex]\begin{gathered} -1000=40\times a \\ \Rightarrow a=\frac{1000}{40} \\ =-25\text{ m/s}^2 \end{gathered}[/tex]
As the foam balls oppose the motion of the gymnast, the gymnast comes to rest after a certain period of time. Thus, the final velocity of the gymnast is v=0 m/s
From the equation of motion,
[tex]v^2-u^2=2ah[/tex]
Where h is the depth reached by the gymnast in the pit before coming to rest.
On substituting the known values,
[tex]\begin{gathered} 0-7^2=2\times-25\times h \\ \Rightarrow h=\frac{-7^2}{2\times-25} \\ =0.98\text{ m} \end{gathered}[/tex]
Thus the gymnast reaches a maximum depth of 0.98 m.
