Asked by chelsea
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?
Answers
Answered by
drwls
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
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
Answered by
Anonymous
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
(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
Answered by
righerthenu
Quadratic formula is (1/2a)(-b +/- sqrt(b^2-4ac))! the Anonymous was wrong
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