A ball is dropped from a height. If it takes 0.2s to cross the last 6 m before hitting the ground, find the height from which it was dropped.

Height = H
Total time = T

H = 0.5gT^2

H - 6 = 0.5g(T - 0.2)^2
0.5gT^2 - 6 = 0.5g(T^2 + 0.04 - 0.4T)
0.5gT^2 - 6 = 0.5gT^2 + 0.02g - 0.2gT
-6 = 0.02g - 0.2gT
T = (0.02g + 6) / (0.2g)
T = [(0.02 * 10) + 6) / (0.2 * 10)
T = 3.1 s

H = 0.5gT^2
= 0.5 * 10 * 3.1^2
= 48.05 m

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