A stone is dropped off a cliff, the last 1.5 seconds before it reaches the ground it travels 30 meters. Find the height from which it dropped & final velocity

Answer 1

distance during that
time=v'*1.5 -4.9*1.5^2
-30=v'*1.5-11.025
v'=-20+7.25=-14.75 m/s
WEll, you know that in each second and a half, the velocity increases by 9.8*1.5 m/s
so v' must represent the 14.75/(9.8*1.5)=.5 second after dropping, so it was dropped 2 seconds before it hit the ground, and it fell...
d=1/2 g t^2

check this, I did most of it in my head.

Answer 2

V'= -12.75m/s not -14.75 m/s

14.75/(9.8*1.5)≠.5