A rock dropped from a cliff falls one-third of its total distance to the ground in the last second of its fall. How high is the cliff?
The answer is 145.7m. But I don't know how to get this answer. Can someone explain it to me?
Thanks.
The answer is 145.7m. But I don't know how to get this answer. Can someone explain it to me?
Thanks.
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Okay, here's my answer.
(It may seem long, but I'm doing my best to explain it carefully.)
Let's say that the height of the cliff is h, and that the rock begins its final second at moment t.
The distance traveled in the final second is equal to:
h/3 = 4.9 + 9.8t (eq.1)
The distance traveled from the top of the cliff to the beginning of the final second is equal to:
2h/3 = 4.9t^2 (eq.2)
Multiplying eq.1 by 2, we get:
2h/3 = 9.8 + 19.6t (eq.3)
Combining eqs. 2 and 3, we get:
4.9t^2 = 9.8 + 19.6t (eq.4)
Moving all of the terms to the left side, we have:
4.9t^2 - 19.6t - 9.8 = 0 (eq.5)
Applying the quadratic equation, we have:
t = {19.6 +/- sqrt(19.6^2 - 4*4.9*-9.8)}/(2*4.9)
...which produces two answers for t:
t = 4.4495 seconds, and, t = -0.4495 seconds.
Only one of these roots makes sense - t = 4.4495 seconds.
So, let's check this answer.
The rock falls for a total of 5.4495 seconds.
4.4495 + 1
During this interval, the distance traveled is:
4.9 * 5.4495^2 = 145.515 meters.
During the first 4.4495 seconds, the distance traveled equals:
4.9 * 4.4495^2 = 97.010 meters.
The rock travels 48.505 meters in the final second - 1/3rd of the total distance.
(It may seem long, but I'm doing my best to explain it carefully.)
Let's say that the height of the cliff is h, and that the rock begins its final second at moment t.
The distance traveled in the final second is equal to:
h/3 = 4.9 + 9.8t (eq.1)
The distance traveled from the top of the cliff to the beginning of the final second is equal to:
2h/3 = 4.9t^2 (eq.2)
Multiplying eq.1 by 2, we get:
2h/3 = 9.8 + 19.6t (eq.3)
Combining eqs. 2 and 3, we get:
4.9t^2 = 9.8 + 19.6t (eq.4)
Moving all of the terms to the left side, we have:
4.9t^2 - 19.6t - 9.8 = 0 (eq.5)
Applying the quadratic equation, we have:
t = {19.6 +/- sqrt(19.6^2 - 4*4.9*-9.8)}/(2*4.9)
...which produces two answers for t:
t = 4.4495 seconds, and, t = -0.4495 seconds.
Only one of these roots makes sense - t = 4.4495 seconds.
So, let's check this answer.
The rock falls for a total of 5.4495 seconds.
4.4495 + 1
During this interval, the distance traveled is:
4.9 * 5.4495^2 = 145.515 meters.
During the first 4.4495 seconds, the distance traveled equals:
4.9 * 4.4495^2 = 97.010 meters.
The rock travels 48.505 meters in the final second - 1/3rd of the total distance.