One-Dimensional Kinematics Problem

In summary: The last one was incorrect. In summary, a physicist notices a ball moving straight upward just outside her window. The ball is visible for an amount of time 0.291 s. The physicist calculates the average velocity as 4.19 m/s and then determines the velocity at the bottom of the window as 7.05 m/s. After that, the physicist tries to use the formula to find the position as a function of time with the formula x= x_o + v_0t - .5gt^2. However, the physicist is not able to solve for x and is not sure how
  • #1
UCrazyBeautifulU
33
0
Hey guys,

I just joined and I have a question. Here is my problem:

Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible 0.291 s as it moves a distance of 1.22 m from the bottom to the top of the window. What is the greatest height of the ball above the top of the window?

How long does it take before the ball reappears?

This is what I have done so far & I have worked on this problem for days and I am ready to cry, seriously.

I figured out the average velocity as 4.19 m/s.

Then I figured out the velocity at the bottom of the window as 7.05 m/s. After that I try to use the formula to find the position as a function of time with the formula x= x_o + v_0t - .5gt^2. At this point I don't know what to do. Do I put in 0 for x_0 or do I put in 1.22? If I put in 0 and use 0.291 as the time I end up with x = 1.64 m. THat answer isn't right according to LON CAPA. ANd after I do figure out the position how do i figure out the time. Do I just divide the position by the 7.05? I need some major help. I need to know how to set up the last formula. Maybe I need to add my answer to the 1.22 or subtract it. I really don't know. Let me know if anyone can help me. Thank you.

You can also send me an email if you don't want to post on the board.

jackie
 
Last edited:
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  • #2
We must determine the velocity at the bottom of the window, v0, and the velocity at the top of the window, v.

For two unknowns we need two equations:

(v + v0)/2 = average velocity

v2 = v02 + 2ax

These hints may provide a good starting help for this problem. The other calculations are more accessible.

UCrazyBeautifulU said:
Then I figured out the velocity at the bottom of the window as 7.05 m/s.
Not correct.
 
  • #3
is the velocity at the bottom of the window 5.62 m/s and the velocity at the top of the window 2.76 m/s? Was the 4.19 m/s correct as the average velocity? Do i use 1.22 as x_0 in the position equation or do I just use 0?

if i use 1.22 i get x=2.44, if i use 0 i get x=1.22. Are either of those correct as the greatest height? And if they are how do I figure out the time after that?
 
  • #4
anyone else have any advice on this one? the above advice didnt really help me. I need some step by step instructions or something. i have been working on this problem for too long.
 
  • #5
Using the 2 equations PPonte gave you, find the velocity at the top of the window. Then use this in the equation v2 = v02 + 2ax with v set to 0, as it is stationary at its greatest height, and v0 the velocity at the top of the window, and solve for x.

Then you need t for traveling a distance 2x, so use the equation d=ut+0.5at2, and solve for t.
 
  • #6
I'm not eaxtly sure if this would work but you could use this equation to try and solve for time portion of your problem; V^2=Vi^2+2ax. You know the initial and final velocity and you the displacement is 1.22m. I would sove for acceleration and then use this equation; (Vf-Vi)/a=t. I also think your initial position is zero and the ball moves a total of 1.22m. Hopefully this will help.
 
  • #7
So I figured out the first answer. It is 0.391 m. Now I need to figure out the time. When i put it in the equation i get two answers...0.286 and 0.278...my equation is

0.391 = 2.77t - .5(9.81)t^2

What is the correct answer for the second part of the problem originally posted in the beginning?
 
  • #8
i tried 0.286 and it was incorrect and i only have one more try left.
 
  • #9
is it 0.572? that is twice 0.286
 
  • #10
UCrazyBeautifulU said:
i tried 0.286 and it was incorrect and i only have one more try left.
One more try left?
What are you referring to?

If this is some homework assignment and you really have trouble understanding this even after the generous hints it is not going to help you by just getting the answers here and giving the teacher the right numbers. That way you are fooling the teacher and yourself.
 
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  • #11
I have gotten hints from the teacher also. we have emailed back and forth. Before I did get help from him I tried a lot of numbers I came up with on LON CAPA, therefore, I only have one more try left, so I want to make sure I am doing it the right way. I know I am close and I have the right formula. I have been working on this for a while.
 

What is one-dimensional kinematics?

One-dimensional kinematics is a branch of physics that studies the motion of objects along a single dimension, typically represented by a straight line. It involves analyzing the position, velocity, and acceleration of an object in relation to time.

What are the basic equations used in one-dimensional kinematics?

The three basic equations used in one-dimensional kinematics are:

  • Position equation: x = x0 + v0t + 1/2at2
  • Velocity equation: v = v0 + at
  • Acceleration equation: v2 = v02 + 2a(x-x0)

How do you solve one-dimensional kinematics problems?

To solve one-dimensional kinematics problems, you need to identify the known and unknown variables, choose the appropriate equation, and plug in the values. It is important to pay attention to units and use the correct sign conventions for direction. You may also need to use multiple equations and algebraic manipulation to solve for the unknown variable.

What is the difference between distance and displacement in one-dimensional kinematics?

Distance refers to the total length traveled by an object, while displacement refers to the change in position of an object from its initial position to its final position. Distance is a scalar quantity, while displacement is a vector quantity that takes into account direction.

How does one-dimensional kinematics relate to real-life situations?

One-dimensional kinematics has many real-life applications, such as calculating the speed and distance of a moving vehicle, determining the acceleration of objects in free fall, and understanding the motion of projectiles. It also plays a role in designing and analyzing the performance of machines and vehicles.

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