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?
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
Quadratic formula is (1/2a)(-b +/- sqrt(b^2-4ac))! the Anonymous was wrong
To find the height of the cliff, we can use a basic physics equation related to free fall. The distance covered by a falling object can be calculated using the formula:
d = (1/2) * g * t^2
where:
d is the distance covered
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time of fall
Given that the rock falls one-third of its total distance in the last second, we can assume that it completes two-thirds of its fall in the time before the last second.
Let's denote the total distance fallen by "x." Therefore, in the last second, the distance fallen is (1/3) * x. And in the time before the last second, the distance fallen is (2/3) * x.
Using the formula, we can write two equations:
(1/3) * x = (1/2) * g * 1^2 (1)
(2/3) * x = (1/2) * g * (t - 1)^2 (2)
Notice that the time in equation (2) is reduced by 1 second because the last second is excluded.
Now, we can solve these equations to find the value of x, which represents the total distance fallen by the rock.
From equation (1):
(1/3) * x = (1/2) * g * 1^2
(1/3) * x = (1/2) * g
From equation (2):
(2/3) * x = (1/2) * g * (t - 1)^2
Dividing equation (2) by equation (1):
(2/3) * x / [(1/3) * x] = [(1/2) * g * (t - 1)^2] / [(1/2) * g]
2 = (t - 1)^2
Taking the square root of both sides:
√2 = t - 1
Simplifying:
t - 1 = √2
t = √2 + 1
Now, let's substitute the obtained value of t back into equation (1) to find x:
(1/3) * x = (1/2) * g * 1^2
(1/3) * x = (1/2) * g
Substituting the value of g and solving for x:
(1/3) * x = (1/2) * 9.8
(1/3) * x = 4.9
x = 4.9 * 3
x = 14.7 meters
Therefore, the total distance fallen by the rock is 14.7 meters, which represents the height of the cliff.
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