The time it takes for an object dropped from the top of cliff A to hit the water in the lake below is twice the time it takes for another object dropped from the top of cliff B to reach the lake. If it takes 2.30 s for the object to fall from cliff A to the water, what are the heights of cliffs A and B?
For u=initial velocity =0,
H (height for free fall) = (1/2)gt².
Since t is known, solve for H in each case.
Well, let's start by giving Cliff A a bit of an ego boost. Cliff A is clearly the overachiever here, taking half the time to reach the lake compared to Cliff B. It's like Cliff A has no wait(ing) issues.
Now, let's do some math-ing. Since the time it takes for an object to fall from Cliff A is twice the time it takes for an object to fall from Cliff B, we can say that:
Time taken from Cliff B = 2.30 s / 2 = 1.15 s
Now, let's calculate the height of the cliffs using a little bit of physics. The height (h) can be determined using the equation:
h = 0.5 * g * t^2
where g is the acceleration due to gravity (9.8 m/s^2) and t is the time of fall.
For Cliff A:
2.30 s = 0.5 * 9.8 m/s^2 * t^2
t^2 = 2.30 s / (0.5 * 9.8 m/s^2)
t^2 = 0.468 s
t = √(0.468 s)
t ≈ 0.684 s
Now, let's plug Cliff B's time of 1.15 s into the equation:
h = 0.5 * 9.8 m/s^2 * (1.15 s)^2
h = 0.5 * 9.8 m/s^2 * 1.3225 s^2
h ≈ 6.33 m
So, the height of Cliff A is approximately 0.684 m and the height of Cliff B is approximately 6.33 m. Cliff B might be a little late to the party, but at least it's not as low as its self-esteem might suggest!
Let's assume the height of cliff A is represented by hA, and the height of cliff B is represented by hB.
We can use the equation of motion for free fall:
h = (1/2) * g * t^2
where h represents the height, g represents the acceleration due to gravity, and t represents the time.
For the object falling from cliff A, we have:
hA = (1/2) * g * tA^2 ..........(1)
For the object falling from cliff B, we have:
hB = (1/2) * g * tB^2 ..........(2)
Given that the time it takes for the object to fall from cliff A to the water is 2.30 s, we can substitute this into equation (1):
hA = (1/2) * g * (2.30)^2
Simplifying this equation, we get:
hA = 2.3^2 * (1/2) * g
Since the time it takes for the object to fall from cliff A to the water is twice the time for the object falling from cliff B to reach the water, we can write:
tA = 2 * tB
Substituting tA = 2.30 s and rearranging the equation, we get:
tB = tA / 2 = 2.30 / 2 = 1.15 s
Now, substituting tB = 1.15 s into equation (2), we get:
hB = (1/2) * g * (1.15)^2
Therefore, the heights of cliffs A and B are:
hA = 2.3^2 * (1/2) * g
hB = (1/2) * g * (1.15)^2
To find the heights of cliffs A and B, we need to use the kinematic equation:
h = (1/2) * g * t^2
where:
h is the height of the cliff
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken for the object to fall
Let's start by finding the height of cliff A.
Given that it takes 2.30 seconds for the object to fall from cliff A to the water, we can substitute these values into the equation:
2.30 = (1/2) * (9.8) * (t^2)
Simplifying the equation, we have:
2.30 = 4.9 * t^2
Dividing both sides by 4.9 to isolate t^2, we get:
0.4694 = t^2
Taking the square root of both sides, we find:
t ≈ 0.684 seconds
Now that we know the time it takes for the object to fall from cliff B (0.684 seconds), and we are given that this time is half the time it takes for the object from cliff A to reach the water, we can calculate the height of cliff B.
Let's set up the equation:
0.684 = (1/2) * (9.8) * (t^2)
Simplifying further:
0.684 = 4.9 * t^2
Dividing both sides by 4.9, we get:
0.1396 = t^2
Taking the square root again, we find:
t ≈ 0.374 seconds
Now we have the time it takes for the object to fall from cliff B (0.374 seconds). We can plug this value back into the kinematic equation to find the height of cliff B:
h = (1/2) * (9.8) * (t^2)
h = (1/2) * (9.8) * (0.374^2)
h ≈ 0.6909 meters
So, the height of cliff B is approximately 0.6909 meters, and the height of cliff A is 2.30 meters.