A rock is dropped from rest into a well. The sound of the splash is heard 2.40 s after the rock is
released from rest. (a) How far below the top of the well is the surface of the water? The speed of sound
in air (at the ambient temperature) is 336 m/s. (3p) (b) What If? If the travel time for the sound is
neglected, what percentage error is introduced when the depth of the well is calculated?
Speed of sound, s=336 m/s
depth, x m
initial velocity = 0 =u m/s
total time, T = 2.4 seconds
time to descendm, t= 2.4 - x/s
x=ut+(1/2)gt²
t+x/s=2.4
t+(ut+(1/2)gt²)/s=2.4
Solve for t. I get t=2.32(approx.)
x=ut+(1/2)gt²
=26.4 m (approx.)
For part b, calculate x when t=2.4 and compare with part a.
I get 28.2m
Well, well, well, let's calculate! (pun intended)
(a) To find the distance the rock traveled before hitting the water, we need to determine the time it took. The time it took for the sound to reach us is given as 2.40 s.
So, the time it took for the rock to reach the water is the same, 2.40 s.
Now, we can use the equation of motion for free fall to find the distance:
s = (1/2) * g * t^2
where
s is the distance (what we're looking for),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time (2.40 s).
Plugging in these numbers, we get:
s = (1/2) * 9.8 m/s^2 * (2.40 s)^2
Calculating that, we find:
s ≈ 28.14 meters
So, the surface of the water is approximately 28.14 meters below the top of the well.
(b) Ah, the "What If?" game! If we neglect the travel time for the sound, we're assuming that the sound reaches us instantaneously. But in reality, sound takes time to travel through air.
To calculate the percentage error, we need to compare the actual time it takes for the sound to reach us with the calculated time (2.40 s).
Let's assume the actual time it would take for the sound to reach us is t_actual.
The percentage error can be calculated using the formula:
Percentage error = [(t_actual - 2.40 s) / t_actual] * 100
Now, since the speed of sound in air is given as 336 m/s, and the distance from the water surface to the top of the well is approximately 28.14 meters, we can calculate the actual time it took for the sound to reach us:
t_actual = distance / speed
t_actual = 28.14 m / 336 m/s
Calculating that, we find:
t_actual ≈ 0.084 s
Now, let's calculate the percentage error:
Percentage error = [(0.084 s - 2.40 s) / 0.084 s] * 100
Calculating that, we find:
Percentage error ≈ -2,757%
So, neglecting the travel time for the sound around the well would introduce a percentage error of approximately -2,757%.
But hey, don't worry too much about it! In the grand scheme of things, it's just a small drop in the well, right?
To answer these questions, we'll need to use the formulas for distance and time. Let's go step by step:
(a) To find how far below the top of the well the surface of the water is, we need to calculate the distance the rock falls in 2.40 seconds.
We can use the formula for distance traveled by an object in free fall:
d = 1/2 * g * t^2
Where:
d is the distance traveled
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
In this case, the rock is dropped from rest, so its initial velocity is 0.
Plugging in the values:
d = 1/2 * 9.8 m/s^2 * (2.40 s)^2
Calculating this, we get:
d = 1/2 * 9.8 m/s^2 * 5.76 s^2 = 28.224 m
So, the surface of the water is approximately 28.224 meters below the top of the well.
(b) Now, let's consider the error introduced when neglecting the travel time for the sound.
The time it takes for the sound to travel up from the water surface to the top of the well is the same as the time it takes for the sound to travel down from the top of the well to the water surface. Therefore, the total travel time for the sound is 2.40 s x 2 = 4.80 s.
To calculate the percentage error, we need to compare the neglected travel time (2.40 s) to the total travel time (4.80 s).
Let's calculate the error percentage:
Error % = (Total travel time - Neglected travel time) / Neglected travel time x 100
Error % = (4.80 s - 2.40 s) / 2.40 s x 100
Calculating this, we get:
Error % = 2.40 s / 2.40 s x 100 = 100%
Therefore, neglecting the travel time for the sound would introduce a 100% error in the calculation of the depth of the well.