At a height of 15 meters above the surface of a lake, a sound pulse is generated. The echo from the bottom of the lake returns to the point of origin 0.150 s later. The air and water temperatures are 20°C. How deep is the lake?

look up speed of sound in water at 20C

call it Cw
look up speed of sound in air at 20C
call it Ca
then distance in water = 2d
distance in air = 30
time in water = t
time in air = (.15 -t)
then
time in air = 30/Ca = .15 - t
solve for t
then
Cw t = 2 d

To solve this problem, we need to use the speed of sound in both air and water and the time it takes for the echo to return.

Step 1: Find the speed of sound in air.
The speed of sound in air can be calculated using the formula:
v = 331.3 + 0.606 × T
where v is the speed of sound in m/s and T is the temperature in degrees Celsius.

Given that the air temperature is 20°C, we can substitute this value into the formula:
v_air = 331.3 + 0.606 × 20
= 331.3 + 12.12
= 343.42 m/s

So, the speed of sound in air is 343.42 m/s.

Step 2: Find the speed of sound in water.
The speed of sound in water can be calculated using the formula:
v_water = 1481.0 + 4.61 × T - 0.05 × T^2 + 0.0105 × T^3 - 0.000071 × T^4
where v_water is the speed of sound in water in m/s and T is the temperature in degrees Celsius.

Given that the water temperature is also 20°C, we can substitute this value into the formula:
v_water = 1481.0 + 4.61 × 20 - 0.05 × 20^2 + 0.0105 × 20^3 - 0.000071 × 20^4
= 1481.0 + 92.2 - 0.05 × 400 + 0.0105 × 8000 - 0.000071 × 160000
= 1481.0 + 92.2 - 20 + 84 - 11.36
= 1626.84 m/s

So, the speed of sound in water is approximately 1626.84 m/s.

Step 3: Calculate the time it takes for the sound pulse to travel downwards and upwards.
The total time for the echo to return is 0.150 s. Since the sound pulse travels downwards to the bottom of the lake and then returns upwards, we can divide this time by 2 to find the time it takes for each leg of the journey.

t_down = t_up = 0.150 s / 2
= 0.075 s

Step 4: Calculate the depth of the lake.
To calculate the depth of the lake, we can use the formula:
depth = (v_water - v_air) × t_down
where depth is the depth of the lake in meters, v_water is the speed of sound in water, v_air is the speed of sound in air, and t_down is the time it takes for the sound pulse to travel downwards.

Substituting the given values, we have:
depth = (1626.84 m/s - 343.42 m/s) × 0.075 s
= 1283.42 m/s × 0.075 s
= 96.257 m

So, the depth of the lake is approximately 96.257 meters.

To find the depth of the lake, we can use the speed of sound in air and water, along with the time taken for the echo to return.

Step 1: Determine the speed of sound in air and water:
The speed of sound in air can be calculated using the formula:
v_air = 331.4 + 0.6 * T
where T is the temperature in Celsius. Given that the air temperature is 20°C, the speed of sound in air is:
v_air = 331.4 + 0.6 * 20 = 343.4 m/s

The speed of sound in water at a given temperature can be calculated using the formula:
v_water = 1402 + 4.7 * T - 0.055 * T^2 + 0.00029 * T^3
where T is the temperature in Celsius. Given that the water temperature is also 20°C, the speed of sound in water is:
v_water = 1402 + 4.7 * 20 - 0.055 * 20^2 + 0.00029 * 20^3
= 1483 m/s

Step 2: Determine the time taken for the echo to return:
The total time taken for the sound pulse to travel from the surface to the bottom of the lake and back is 0.150 s.

Since the sound travels the same distance both ways, we can divide the total time by 2 to get the time taken for one-way travel.
So, the time for one-way travel is: t = 0.150 s / 2 = 0.075 s

Step 3: Calculate the depth of the lake:
Let's assume the depth of the lake is h.

For the sound pulse to travel down from the surface to the bottom of the lake, it has to cover a distance of 15 meters in time t:
Distance = speed * time
15 = v_air * t

For the echo to travel back from the bottom to the surface, it has to cover a distance of 15 + h in time t:
Distance = speed * time
15 + h = v_water * t

Substituting the values we calculated earlier, we have two equations:
15 = 343.4 * 0.075
15 + h = 1483 * 0.075

Solving these equations will give us the value of h, which is the depth of the lake.