Aroller coaster glides from rest from the top of an 80.0 meter hill. what is the speed of the roller coaster at the bottom of the hill?

So this is the case of energy conversion. From potential to kinetic.
As energy is conserve:
Epi+Eki=Epf+Ekf ; i for initial and f for final. At begining object at rest hence Eki=0 while at final, object on the lowest position, hence epf=0 therefore:
Epi=Ekf
M.g.h=1/2.m.v^2
Assume no mass change during the process, then
G.h=1/2.v^2
Hence v at bottom:
v=sqrt(2×g×h) assume g=10m/s^2
=sqrt(2×10×80)=sqrt(1600)=40m/s
Hence velocity in the bottom is 40 m/s

Answer8.0m;