A rock is dropped into an abandoned mine and a splash is heard 2 s later. Assuming that it takes a negligible time for the sound to travel up the mine shaft, determine the depth of the shaft and how fast the rock was falling when it hit the water.
In other words how far did it fall in 2 seconds
I assume you are using the metric system and g = 9.81 m/s^2
h = (1/2) g t ^2 = 4.9 t^2 = 4.9 * 4
v = g t = 9.81 * 2 meters/second
in general:
if a = -9.81
v = Vi - 9.81 t
y = Yi + Vi t - 4.9 t^2
here y is 0, ground
Vi is 0, initial velocity
Yi is height, what we want to know
To determine the depth of the shaft and the speed at which the rock was falling when it hit the water, we can use the equations of motion.
Let's define the following variables:
- d: Depth of the shaft (unknown)
- g: Acceleration due to gravity (approximately 9.8 m/s^2)
- t: Time taken for the rock to fall (2 s)
- v: Final velocity of the rock when it hits the water (unknown)
Using the equation of motion in free fall, which relates distance (d), initial velocity (u), time taken (t), and acceleration (g):
d = ut + (1/2)gt^2
Since the rock is dropped, the initial velocity (u) is 0, so the equation simplifies to:
d = (1/2)gt^2
Substituting the known values, we have:
d = (1/2)(9.8 m/s^2)(2 s)^2
d = (1/2)(9.8 m/s^2)(4 s^2)
d = 19.6 m
Therefore, the depth of the shaft is 19.6 meters.
To determine the speed of the rock when it hits the water, we can use another equation of motion:
v = u + gt
Since the final velocity (v) is what we want to find, we can rearrange the equation to solve for it:
v = gt
Substituting the known values, we have:
v = (9.8 m/s^2)(2 s)
v = 19.6 m/s
Therefore, the rock was falling at a speed of 19.6 meters per second when it hit the water.
To determine the depth of the shaft and the speed of the falling rock, we can use the equations of motion.
Let's analyze the situation:
1. The rock is dropped, which means its initial velocity (u) is 0 m/s.
2. The only force acting on the rock is gravity, causing it to accelerate downward. We'll use the acceleration due to gravity (g) as 9.8 m/s^2.
3. The time it takes for the splash sound to reach the surface is given as 2 seconds.
First, let's find the depth of the shaft using the formula for the displacement of a falling object:
s = ut + (1/2)gt^2
Since the initial velocity (u) is 0, the equation simplifies to:
s = (1/2)gt^2
Substituting g = 9.8 m/s^2 and t = 2 s:
s = (1/2)(9.8 m/s^2)(2 s)^2
s = (1/2)(9.8 m/s^2)(4 s^2)
s = (1/2)(9.8 m/s^2)(16)
s = 78.4 m
Therefore, the depth of the shaft is 78.4 meters.
Next, let's determine the speed of the rock when it hits the water. We can use the formula:
v = u + gt
Substituting u = 0 m/s and t = 2 s:
v = (0 m/s) + (9.8 m/s^2)(2 s)
v = 0 m/s + 19.6 m/s
v = 19.6 m/s
Therefore, the rock was falling at a speed of 19.6 m/s when it hit the water.