A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.
The well is 80ft deep
There is a special place in Hell for physics students who drop stones in wells. Wells don't need stones dropped in them, or the water level below bottom will decrease.
Let t be the time of dropping the first stone.
disance apart= distance first-distance2nd
= 1/2 g t^2 - 1/2 g(t-1)^2
so, at t=2
= 1/2 g 4 - 1/2 g 1
To find the distance between stones A and B another second later, we first need to determine the distance each stone has traveled in 1 second.
For stone A:
We can use the equation for the 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, and t is the time.
In this case, t = 1 second and g is approximately 32.2 ft/s^2.
So, the distance traveled by stone A in 1 second is:
d_a = (1/2) * 32.2 * (1^2) = 16.1 ft.
For stone B:
Since stone B is dropped 1 second after stone A, it will also fall for 1 second.
Therefore, the distance traveled by stone B in 1 second is also 16.1 ft.
Finally, to find the distance between the stones another second later:
The distance traveled by stone A in 2 seconds is 2 * d_a = 2 * 16.1 = 32.2 ft.
The distance traveled by stone B in 2 seconds is 2 * 16.1 = 32.2 ft.
Since both stones have traveled the same distance, the distance between them another second later is 0 ft (they will be at the same height).
To determine the distance between the two stones another second later, we need to calculate the positions of each stone after 2 seconds.
First, let's calculate the distance traveled by stone A after 2 seconds:
We know that the distance traveled by an object in free fall is given by the formula:
d = (1/2) * g * t^2
Where:
d is the distance traveled
g is the acceleration due to gravity (which is approximately 32 ft/s^2)
t is the time in seconds
For stone A, after 2 seconds, we have:
dA = (1/2) * g * (2^2)
= (1/2) * 32 * 4
= 64 ft
So, after 2 seconds, stone A will have traveled 64 ft.
Now, let's calculate the distance traveled by stone B after 2 seconds:
Since stone B is dropped 1 second after stone A, it will have been falling for 1 second less when stone A has fallen for 2 seconds. Therefore, the time for stone B will be 1 second less, i.e., 2 - 1 = 1 second.
Using the same formula as before, the distance traveled by stone B after 1 second will be:
dB = (1/2) * g * (1^2)
= (1/2) * 32 * 1
= 16 ft
So, after 2 seconds, stone B will have traveled 16 ft.
To determine the distance between the stones another second later, we subtract the distance traveled by stone B from the depth of the well, which is 80 ft:
Distance between the stones = Depth of the well - Distance traveled by stone B after 2 seconds
= 80 ft - 16 ft
= 64 ft
Therefore, another second later, the distance between the stones will be 64 ft.