Calculating Drop Locations: The Physics of Water Dripping from a Shower Nozzle

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The discussion revolves around calculating the drop locations of water from a shower nozzle that is 81 inches above the floor. To find the position of each drop, the position function y = v_{y0}t + 1/2 a_{y}t^2 is suggested, noting that the initial velocity is zero. The timing of the drops is crucial, as the first drop hits the floor exactly when the fourth drop begins to fall, indicating that the drops fall at regular intervals of time, specifically 1/3 of the time it takes for the first drop to reach the floor. This understanding clarifies the timing for when each successive drop starts to fall. The discussion emphasizes the importance of timing in determining the location of the drops upon impact.
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Water drips from the nozzle of a shower onto the stall floor 81 in. below. The drops fall at regular intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the location of the individual drops when a drop strikes the floor.

So we are given a distance the drop falls 81 inches . When it says to find the location of the individual drops, I need a position function. Should I use y = v_{y}_{0}t + \frac{1}{2}a_{y}t^{2}?

Thanks
 
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Right, but there's no initial velocity. And first you need to find the time each drop starts falling.
 
Or, more correctly, the initial velocity is 0! Calculate the time it takes the first drop to hit the floor. According to the problem, that is exactly the time the fourth drop starts to fall. So, since the drops fall "at regular intervals of time", that "regular intervals of time" is 1/3 the time the first drop takes to hit the floor (do you see why it is 1/3 and not 1/4?). Now you know when each successive drop starts!
 
yes that's what i was thinking. its 1/3 and not 1/4, because the first drop hits the ground only when the fourth drop starts to fall, not when it hits the ground.
 
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