Help with Problem (Complicated Kinematics)

  • Thread starter Thread starter AuraofMana
  • Start date Start date
  • Tags Tags
    Kinematics
AI Thread Summary
The discussion revolves around solving a kinematics problem involving a flowerpot that moves up and down past a window. Participants suggest breaking the problem into steps, starting with calculating the average speed of the flowerpot as it passes the window. They emphasize that the velocities at the same height during ascent and descent are equal but in opposite directions, allowing for the use of kinematic equations. Additionally, some participants express humorous skepticism about the scenario's realism, questioning the cat's ability to measure time and communicate findings. The conversation highlights the importance of understanding motion dynamics in kinematics problems.
AuraofMana
Messages
1
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
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.
 
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

v^2 = u^2 - 2gy

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

v = u - gt

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

v_b = v_a - gt

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

v_a = v_b + gt

From this info you can calculate va from

y = v_at - 4.9t^2

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?
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top