A man drops a stone into a water well on his farm. He hears the sound of the splash 3.25 s later. The acceleration of gravity is 9.8 m/s2 and the speed of sound in air is 326 m/s. How deep is the well?
Denote the height by h, and the time for stone to descend t, we have the equation
h=ut+(1/2)gt&sp2;
where
u=0 m/s, initiial speed, and
g=acceleration due to gravity.
For the sound to return, we have
h=326 m/s*(3.25-t)
Solve for h and t using the two equations.
You should have t=3.1 and h=47 m approximately
h=ut+(1/2)gt²
Thanks a lot
How exactly was h and t solved for?
Explain further pls
To find the depth of the well, we need to consider the time it takes for the sound to reach the man after the stone is dropped.
First, let's calculate the time it takes for the stone to reach the bottom of the well. We can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (which is 0, as the stone is dropped from rest)
t = time taken
a = acceleration
In this case, we can rearrange the equation to solve for time:
t = sqrt(2s/a)
Substituting the given values, the time it takes for the stone to reach the bottom of the well (the depth of the well) is:
t = sqrt(2s/9.8)
Next, we need to calculate the time it takes for the sound to reach the man. We know that the speed of sound in air is 326 m/s, and the distance traveled by the sound is equal to the depth of the well. Therefore, the time taken by the sound is:
t = s / 326
Now, let's consider the total time taken. We know that the stone reaches the bottom of the well after 3.25 seconds and that the total time is the sum of the time taken by the stone and the time taken by the sound:
3.25 = sqrt(2s/9.8) + s/326
This equation can be solved to find the value of s, which represents the depth of the well.
100.3 meters, if I'm not mistaken..
Well deep = 326/3.25= ~100m