A certain freely failing object requires 1.70 s to travel the last 20.5 m before it hits the ground. From what height above the ground did it fall?

Answer 1

Let H be the initial height.
Let T be the time it takes to fall the full height H.
g/2 T^2 = 4.9 T^2 = H
(g/2)(T-1.7)^2 = 4.9(T-1.7)^2 = H - 20.5

Combine the equations to eliminate H, so you can solve for T. Then calculate H.

4.9T^2 = 4.9(T-1.7)^2 + 20.5
4.9T^2 = 4.9T^2 -16.66 T +14.161 +20.5
16.66T = 34.66
T = 2.08 s

H = 4.9 T^2 = 21.2 m

Answer 2

Find the height from which you would have to drop a ball so that it would have a speed of 6.4 m/s just before it hits the ground.