Solve Kinematics Problem Involving Free Fall Acceleration

In summary: 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?The second and third drops are released at t=1.125m and t=2.00m after the first drop, respectively.
  • #1
ubiquinone
43
0
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 [tex]2m[/tex] 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: [tex]-2.00m=-\frac{1}{2}(9.8m/s^2)t^2[/tex]
The time was [tex]t=\sqrt{\frac{2.00m}{4.9m/s^2}}\approx 0.64s[/tex]

Now how can I use this information to solve this problem?
 
Last edited:
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  • #2
Firstly, your time is not correct. You have an error in your manipulation; the correct expression is as follows;

[tex]x=\frac{1}{2}at^2 \Leftrightarrow t = \sqrt{\frac{2x}{a}}[/tex]

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?
 
  • #3
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.
 
  • #4
So should I find the how many seconds for one drop to be produce, i.e. [tex]\sqrt{\frac{2x}{g}}\times\frac{1}{4}[/tex]

Then to find the position of the second drop plug into [tex]x=\frac{1}{2}gt^2[/tex], [tex]t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s[/tex]?
 
  • #5
Office_Shredder said:
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: .
 
  • #6
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?
 
  • #7
ubiquinone said:
So should I find the how many seconds for one drop to be produce, i.e. [tex]\sqrt{\frac{2x}{g}}\times\frac{1}{4}[/tex]

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

[tex]t=\frac{2}{3}\sqrt{\frac{2}{4.9}}s[/tex]

into

[tex]x=\frac{1}{2}gt^2[/tex]

To find the distance traveled.
 
Last edited:
  • #8
Okay, then [tex]t=\frac{3}{4}\sqrt{\frac{2}{4.9}}s\approx 0.48s[/tex]
Plugging into, [tex]x=\frac{1}{2}gt^2=\frac{1}{2}(9.8m/s^2)(.48s)=1.125m[/tex]

Is this correct?
 
  • #9
No...
I've done it wrong, I think. I should be multiplying [tex]\frac{2}{3}[/tex] to the time instead of [tex]\frac{3}{4}[/tex] otherwise, it would mean once a drop starts to drip, the 4th drop below it had hit the ground 2.00 m below.
 
  • #10
2/3 is correct since it starts to fall at t=1/3 (.64)
so now it's not too hard.
 

Related to Solve Kinematics Problem Involving Free Fall Acceleration

1. What is free fall acceleration?

Free fall acceleration is the acceleration of an object due to the force of gravity. It is a constant value of 9.8 meters per second squared (m/s²) on Earth, and it causes objects to accelerate towards the ground with increasing speed.

2. How do you calculate free fall acceleration?

To calculate free fall acceleration, you can use the formula a = g, where "a" represents the acceleration and "g" represents the gravitational constant of 9.8 m/s². Alternatively, you can use the formula a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time.

3. What is the difference between free fall acceleration and normal acceleration?

The main difference between free fall acceleration and normal acceleration is that free fall acceleration only occurs when an object is falling under the influence of gravity, while normal acceleration can occur in any situation where an object's speed or direction is changing.

4. How does air resistance affect free fall acceleration?

Air resistance, also known as drag, can slow down the acceleration of an object in free fall. This is because air resistance creates an opposing force that acts against the force of gravity, causing the object to experience a lower net force and therefore a lower acceleration.

5. What are some real-life examples of free fall acceleration?

Some common examples of free fall acceleration in real life include objects falling from a height (such as a skydiver or a dropped pencil), a rollercoaster going down a steep drop, and a ball being thrown into the air and then falling back down. Essentially, any situation where an object is falling under the force of gravity is an example of free fall acceleration.

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