Help with Problem (Complicated Kinematics)

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In summary, the cat observed a flowerpot moving up and down and measured its height above the window. He found that the flowerpot went up and down at the same height, but moved in opposite directions. He used this information to calculate the velocity of the flowerpot.
  • #1
AuraofMana
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Help with Problem please (Complicated Kinematics)

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.35 s, and the top-to-bottom height of the window is 1.95 m. How high above the window top did the flowerpot go?

-At first I tried finding the velocity by m/s. Then later when I thought about it, the velocity is not constant in this problem. Now I have no clue what to do, even though I know 9.8 m/s^2 as gravity and the velocity in the distances of the window.
 
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  • #2
You'll need to attack this problem in steps. There are several ways to go. Try figuring out the speed of the flowerpot at various points. Hint: Start by finding the average speed of the flowerpot as it passes by the window.
 
  • #3
Think about the problem. Will the velocities up and down be the same? If so, then you can calc the initial and final velocities of the pot going past the window, and the rest is easy peasy. Try drawing a diagram also.
 
  • #4
The up and down speeds of the pot is the same at the same height above the ground, they are just in opposite directions. This can be seen from the equation

[tex]v^2 = u^2 - 2gy[/tex]

which do not distinguish between the up and down motions. This means that the up and down times to cross the window will be the same when one considers

[tex]v = u - gt[/tex]

for the separate cases of the up and down motion past the window. For instance say the velocity at the bottom is va and the top vb, then for the going up case (taking up as positive) we have

[tex]v_b = v_a - gt[/tex]

and for the going down case (taking down as positive) we have

[tex]v_a = v_b + gt[/tex]

From this info you can calculate va from

[tex]y = v_at - 4.9t^2[/tex]

But I must say I see several logical errors in the problem:
1. Why would someone want to throw a flower pot upwards? He would just have to catch it again on its way down or jump out of the way.
2. How or why would the cat measure the time for the pot to cross the window?
3. Who taught him to add?
4. How did he communicate his results to the physicist?
5. If the cat did not do it, why is he mentioned in the problem anyway?
 
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Related to Help with Problem (Complicated Kinematics)

1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It involves analyzing the position, velocity, and acceleration of an object.

2. What is complicated kinematics?

Complicated kinematics refers to situations where the motion of an object is not easily described by simple equations or formulas. This could be due to factors such as changing acceleration, non-uniform motion, or multiple objects interacting with each other.

3. How can kinematics be used to solve problems?

Kinematics can be used to solve problems by using mathematical equations and principles to analyze the motion of objects. This can help predict the future position or velocity of an object, or determine the initial conditions required to achieve a desired motion.

4. What are some common equations used in kinematics?

Some common equations used in kinematics include the equations for position (x = x0 + v0t + 1/2at^2), velocity (v = v0 + at), and acceleration (a = (vf - v0)/t), where x is position, v is velocity, a is acceleration, t is time, and subscripts 0 and f represent initial and final values.

5. What are some practical applications of kinematics?

Kinematics has numerous practical applications in fields such as engineering, robotics, and sports. It is used to design and analyze the motion of machines and structures, program movements in robots, and improve athletic performance by understanding the motion of the human body.

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