Answer:
F_A =5.625*10^1^6N
Step-by-step explanation:
From the question we are told that
50,000 years ago,
A giant 4.5 107-kg meteor
180-m-deep hole
20,000 m/s
Generally for this problem the energy change is given as
\triangle E=\frac{1}{2} mv^2 +mgd△E=
2
1
mv
2
+mgd
Having the potential and kinetic energy in place
Mathematically solving for Average forceF_AF
A
\triangle E=F_a*d△E=F
a
∗d
F_A =\frac{1/2* 4.5*10^7(20,000)^2-kg+4.5*10^7*9.81*160}{160}F
A
=
160
1/2∗4.5∗10
7
(20,000)
2
−kg+4.5∗10
7
∗9.81∗160
Therefore Average force F_AF
A
is given by
F_A =5.625*10^1^6N