Unraveling the Flying Leap of a Flea

  • Thread starter Thread starter MozAngeles
  • Start date Start date
  • Tags Tags
    Flying
AI Thread Summary
The discussion revolves around analyzing high-speed motion pictures of a flea's jump to determine its maximum height and acceleration. Participants share methods for calculating the flea's maximum height at 2.5 ms, with one user estimating it to be about 0.25 cm using the area under the graph rather than traditional kinematics equations. They suggest that similar calculations can be applied to find the flea's height at 0.5 ms, 1.0 ms, and 1.5 ms by evaluating the areas of triangles and rectangles on the graph. The conversation emphasizes using graphical data for accurate estimations in physics problems. Overall, the thread highlights the importance of visual data interpretation in solving kinematic questions.
MozAngeles
Messages
100
Reaction score
0

Homework Statement


High-speed motion pictures 3500 (frames/second) of a jumping, 210 μg flea yielded the data used to plot the graph given in the figure . (See "The Flying Leap of the Flea" by M. Rothschild, Y. Schlein, K. Parker, C. Neville, and S. Sternberg in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical take-off angle. Use the graph to answer the questions.
YF-02-42.jpg


Find the maximum height the flea reached in the first 2.5 {\rm ms}.
Find the flea's acceleration at 0.5 {\rm ms}.

Homework Equations



the kinematics equations
x= .5at2+ v0+ x0
v= at + v0
...

The Attempt at a Solution


hmax=1/2g*t
after converting everything to the right units i got .0031 cm which was wrong... help meh please?
 
Physics news on Phys.org
how do you know it was wrong? do you have the answer? I've done a quick estimate giving 0.25ish cm. if that's closer I'll explain how I did it.
 
Last edited:
yeah that's the correct answer! :) how did you do it?
 
Well... I didn't use any formulae in my estimation really, I just used the fact that the distance traveled = the area under the graph...

if you do it in two halves (triangular and rectangular)
you get height (distance traveled upwards buy flea) = (1/2)*(1/1000 sec)*(125 cm/sec) + ((2.5-1)/1000 sec)*(125 cm/sec) = about 0.25cm

this is actually a fair approximation as the extra bit added to the rectangle at the top left corner by taking the left hand edge as 1ms compensates for the bit missing from the triangle caused by the fact that the slope upwards in concave

There may be a more "mathsy" way of doing this, perhaps more accurately, but the question does after all say "Use the graph to answer the questions."

Hope this helps!

Will
 
could you use the same concept to find the flea's height at 0.5 ms, 1.0, and 1.5 ms??
 
I don't see why not? just work out the area under the graph in terms of triangles and rectangles up to those times on the x-axis.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...

Similar threads

Back
Top