Can You Calculate the Maximum Distance in an Egg Toss Without Initial Velocity?

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SUMMARY

The discussion centers on calculating the maximum distance an egg can be tossed without breaking, given a breaking force of 5N and an egg mass of 50g. Participants estimate that the maximum acceleration the egg can withstand is 100 m/s², leading to a proposed maximum distance of 100 meters. The conversation highlights the importance of initial velocity and centripetal acceleration, with one user calculating a velocity of 31.6 m/s based on a 0.5 m arm length and a launch angle of 60 degrees. The need for precise calculations and assumptions is emphasized throughout the discussion.

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two people toss a raw egg in the egg toss. the force required to break the shell is 5N. the mass of the egg is 50g. estimate the maximum separation distance for the egg throwers. make whatever reasonable assumptions you need to about the launch angle and hand movement.

isnt an initial velocity needed for this problem?
 
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Estimate!

:smile:
 
Originally posted by formulajoe
isnt an initial velocity needed for this problem?
That's what you've got to figure out. How fast can you toss that egg without breaking it?
 
Suppose the tossers are very, very, very strong. Then what?
 
Originally posted by gnome
Suppose the tossers are very, very, very strong. Then what?
Hint: How strong do they have to be to exert 5 N?
 
okay, so i figured the acceleration for the egg to break to be 100 m/s^2.
im estimating the max distance to be 100 m.
im stuck. how do i figure in the 100 m/s^2? the acceleration in both directions is going to be constant. 0 in the x direction and 9.8 in the y direction.
so if the tossers are very strong, they would throw the egg very hard. leaving a high initial velocity. but that doesn't help with how to get the acceleration so high.
 
How long is your arm?
 
oooohhh, use centripetal acceleration to get initial velocity?
 
Originally posted by formulajoe
okay, so i figured the acceleration for the egg to break to be 100 m/s^2.
im estimating the max distance to be 100 m.
The max distance is what you're supposed to be figuring out. Don't guess at that answer!

You figured out the max acceleration the egg can withstand. Now use Bystander's hint!
 
  • #10
using centripetal acc. i found the velocity to be 31.6 m/s with a radius(arm length) of .5 m. I am assuming the launch angle to be 60 deg from the horizontal.
i have a feeling I am way off though.
am i getting close?
 
  • #11
Trout or Carp

You sound like you are from a town named after a fish.
 
  • #12
t

Tom, are you still around. I looked at it like Work = mV^2 where V is the horizontal velocity. I don't know if I am right.
 
  • #13
uhhh how do you know my name?
 
  • #14
?

Because you told me about this place at church today.
 

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