How to calculate the location of drops in a falling water shower?

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SUMMARY

The discussion focuses on calculating the positions of water drops from a shower nozzle as they fall to the floor 215 cm below. The first drop strikes the floor as the fourth drop begins to fall, indicating that the drops fall at regular intervals of time. The key equations used include the kinematic equation \( x(t) = X_0 + \frac{1}{2} a t^2 \) with acceleration due to gravity set at 980 cm/s². The correct approach involves determining the time each drop has been falling when the first drop reaches the ground, allowing for accurate calculations of the positions of the second and third drops.

PREREQUISITES
  • Understanding of kinematic equations, specifically \( x(t) = X_0 + \frac{1}{2} a t^2 \)
  • Basic knowledge of gravitational acceleration (980 cm/s²)
  • Concept of regular time intervals in motion
  • Ability to manipulate algebraic equations for solving time and distance
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn about free fall motion and its characteristics
  • Explore the concept of uniform acceleration and its implications in real-world scenarios
  • Practice solving similar problems involving multiple objects in free fall
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of kinematic equations in practical applications.

way2fasts
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Homework Statement


Water drips from the nozzle of a shower onto the floor 215 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall.

a) Find the location of the third drop as measured from the nozzle of the shower when the first drop strikes the floor. in cm below nozel

b) Find the location of the second drop as measured from the nozzle of the shower when the first drop strikes the floor.


Homework Equations





The Attempt at a Solution




Ok I am going to be honest never have posted here, but I am really confused about this problem, hopefully someone is able to explain this to me.

Well here is what I did, I thought that because there were 4 drops of water i would make t=3t.

then with the x(t)= Xo + 1/2(a)t^2 formula to find out the total time.
0=215+(1/2)(980)t^2
-215=-490t^2
-215/490=t^2
t=.662401

I thought that by getting the total time I would be able to just divide that by 3 to get just T alone so then I would only have to plug the time at that certain interval into a velocity equation. But the problem is when i do this my position is fairly high for 2 and 3, its about 215cm for 2, and 329 cm away from the nozel at 3. can anyone help to show me what I am doing wrong
 
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way2fasts said:
then with the x(t)= Xo + 1/2(a)t^2 formula to find out the total time.
0=215+(1/2)(980)t^2
-215=-490t^2
-215/490=t^2
t=.662401

I thought that by getting the total time I would be able to just divide that by 3 to get just T alone so then I would only have to plug the time at that certain interval into a velocity equation.

Although you handle the -signs in a random way, the result for the time is correct. It is also true that the drops follow each other at t/3 seconds.

But what do you mean on plugging into a velocity equation? You need the distance of the drops from the nozzle. The third drop had traveled for t/3 s when the fourth drop started, that is, when the first one hit the ground. What is its distance travelled?

ehild
 

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