A coin is dropped from rest into a well with water 45.0m below the hand that drops the coin.
A) How much time passes between the release of the coin and its contact with the water?
B)How fast will the coin be going when it lands in the pool?
C)How long after the release of the coin must the dropper wait to actually hear the sound of the water? Take the speed of sound to be v=335 m/s.
How can I find the answer to these questions without using Kinematics?
You have to consider the time going down, and the time for the sound to come up.
Remember the time after the release of the coin will be the sum of these two times. I will be happy to critique your thinking or work.
To solve these questions without using kinematics, we can use basic principles of physics.
A) The time it takes for the coin to reach the water can be calculated using the equation of motion for constant acceleration in the vertical direction:
h = (1/2) * g * t^2
where h is the distance (45.0m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
To find the time, rearrange the equation:
t = sqrt(2h/g)
Substituting the given values:
t = sqrt(2 * 45.0 / 9.8) ≈ 3.03 seconds
Therefore, it will take approximately 3.03 seconds for the coin to reach the water.
B) The final velocity of the coin when it lands in the pool can be calculated using the equation:
v = u + gt
where v is the final velocity, u is the initial velocity (which is zero in this case since the coin is dropped from rest), g is the acceleration due to gravity, and t is the time of fall (calculated in part A).
Substituting the given values:
v = 0 + 9.8 * 3.03 ≈ 29.79 m/s
Therefore, the coin will be going approximately 29.79 m/s when it lands in the pool.
C) To find the time it takes for the sound to reach the surface, we need to consider the distance traveled by the sound wave. The sound wave will travel a distance equal to the total depth of the well (45.0m) twice (down and up).
The time it takes for the sound to travel a given distance can be calculated using the equation:
t = d/v
where t is the time, d is the distance, and v is the speed of sound.
Substituting the given values:
t = (2 * 45.0) / 335 ≈ 0.269 seconds
Therefore, the dropper must wait approximately 0.269 seconds after the release of the coin to actually hear the sound of the water.
Remember to always check your calculations and units to ensure accurate results.
To find the answer to these questions without using kinematics, we can use the concept of free fall and the speed of sound.
A) Time taken for the coin to fall:
We know that the acceleration due to gravity (g) is approximately 9.8 m/s^2. The distance the coin travels is 45.0 m (depth of the well).
Using the equation of motion for free fall:
distance = (1/2) * g * time^2
Rearranging the equation:
time^2 = (2 * distance) / g
Substituting the values:
time^2 = (2 * 45.0 m) / (9.8 m/s^2)
time^2 = 90.0 m / 9.8 m/s^2
time^2 = 9.18 s^2
Taking the square root to find time:
time ≈ √9.18 s^2
time ≈ 3.03 s
So, the time passed between the release of the coin and its contact with the water is approximately 3.03 seconds.
B) The speed of the coin when it lands in the pool:
Using the equation of motion for free fall:
final velocity = initial velocity + (acceleration * time)
Since the coin starts from rest (initial velocity = 0 m/s) and the acceleration is due to gravity (acceleration ≈ 9.8 m/s^2), we can calculate the final velocity:
final velocity = 0 m/s + (9.8 m/s^2 * 3.03 s)
final velocity ≈ 29.75 m/s
So, the coin will be going approximately 29.75 m/s when it lands in the pool.
C) The time taken for the sound to travel up:
The speed of sound is given as v = 335 m/s. The distance the sound needs to travel is 45.0 m.
Using the equation:
time = distance / speed
Substituting the values:
time = 45.0 m / 335 m/s
time ≈ 0.134 s
So, the person must wait approximately 0.134 seconds after releasing the coin to actually hear the sound of the water.
Therefore, the total time after the release of the coin, considering both the time for the coin to fall and the time for the sound to travel up, is approximately 3.03 seconds + 0.134 seconds = 3.164 seconds.