a rock is dropped from a cliff falls one-third of it's total distance to the ground in the last second of it's fall. how high is the cliff?

Answer 1

Let total time to fall be T.
In the last second, the distance covered is
L =(g/2)[T^2 - (T-1)^2]
= 1/3 (g/2) T^2
Cancel the g/2 factors
2T - 1 = (1/3) T^2
T^2 -6T +3 = 0
The roots are
T = (1/2)[6 +/- sqrt(24)] = 0.55 and 5.45 seconds

You cannot use the T = 0.55 solution because it has to fall at least one second.

The height is (g/2)T^2 = 146 meters

Answer 2

remember the quadratic formula

(1/2a)(-b +/- sqrt(4ac)) is helpful when solving for T if you get stuck at T^2-6T+3

where aT^2+bT+c

Answer 3

Quadratic formula is (1/2a)(-b +/- sqrt(b^2-4ac))! the Anonymous was wrong