#color(blue)("at "60"m"#
Here, all the potential energy has NOT been converted into kinetic energy. So we can't use the conservation of (mecahnical) energy formula
We first need to find the time.
#color(red)(d=v_itxx1/2at^2# ----- #2^"nd" " kinematic equation"#
#=>60m=0txx1/2(9.8)(t)^2#
#=>(60cancelmxx2)/(9.8cancelm/s^2)=t^2#
#=>t=sqrt(12.24s^2)~~3.5s#
Now, let's find the #v_f# using #color(red)(v_f=v_i+g t#
#v_f=0+(-9.8m/s^2)(3.5s)#
#=>v_f=color(green)(-34.3m/s#
Now let's find the velocity #color(blue)("at "125"m"#.
Here, all the potential energy has been converted into kinetic energy. So we can use the conservation of (mecahnical) energy formula .
#color(red)(cancelmgh=1/2cancelmv^2#
#color(red)(gh=1/2v^2#
#9.8m/s^2xx125m=1/2 v^2#
#1225m^2/s^2xx2=v^2#
#=>v=sqrt(1225m^2/s^2xx2)=49.49m/s#
#=color(green)(-50m/s# because its in the downward direction.
