Solve Kinematics Problem Involving Free Fall Acceleration

[ubiquinone]

Hi, I have a kinematics problem involving free fall acceleration. I'm not sure how to do this so I was wondering if someone could please give me a hand. Thank you.

Question: Water drips from the nozzle of a shower onto the floor 2m 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. When the first drop strikes the floor, how far belowthe nozzle are the second drop?

I started the problem off by figuring out how long it would take for the first drop to hit the floor: -2.00m=-\frac{1}{2}(9.8m/s^2)t^2
The time was t=\sqrt{\frac{2.00m}{4.9m/s^2}}\approx 0.64s

Now how can I use this information to solve this problem?

 

[Hootenanny]

Firstly, your time is not correct. You have an error in your manipulation; the correct expression is as follows;

x=\frac{1}{2}at^2 \Leftrightarrow t = \sqrt{\frac{2x}{a}}

You know that within the time take for the first drop to hit the floor, the fourth drop has just been produced. What is the time interval between successive drops?

 

[Office_Shredder]

Hoot, the time looks OK to me. 2*2/9.8 = 2/4.9

He just divided the 9.8 by 2 before bringing it to the other side.

 

[ubiquinone]

So should I find the how many seconds for one drop to be produce, i.e. \sqrt{\frac{2x}{g}}\times\frac{1}{4}

Then to find the position of the second drop plug into x=\frac{1}{2}gt^2, t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s?

 

[Hootenanny]

Hoot, the time looks OK to me. 2*2/9.8 = 2/4.9

He just divided the 9.8 by 2 before bringing it to the other side.

Indeed it is, my apologies! I think its time for a break :zzz: .

 

[Ja4Coltrane]

Let t=.64 sec=T

the first drop is released at t=0 sec, the last at t=T
since it is regular, when are the second and third drops released?

 

[Hootenanny]

So should I find the how many seconds for one drop to be produce, i.e. \sqrt{\frac{2x}{g}}\times\frac{1}{4}

Then to find the position of the second drop plug into x=\frac{1}{2}gt^2, t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s?

Yes, so you plug

t=\frac{2}{3}\sqrt{\frac{2}{4.9}}s

into

x=\frac{1}{2}gt^2

To find the distance traveled.

 

[ubiquinone]

Okay, then t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s\approx 0.48s
Plugging into, x=\frac{1}{2}gt^2=\frac{1}{2}(9.8m/s^2)(.48s)=1.125m

Is this correct?

 

[ubiquinone]

No...
I've done it wrong, I think. I should be multiplying \frac{2}{3} to the time instead of \frac{3}{4} otherwise, it would mean once a drop starts to drip, the 4th drop below it had hit the ground 2.00 m below.

 

[Ja4Coltrane]

2/3 is correct since it starts to fall at t=1/3 (.64)
so now it's not too hard.