A 100 metre potato drop will it be unharmed (check)

In summary, the conversation involves the speaker discussing their hypothesis on whether a potato can survive a 100-meter drop. They calculate the force and acceleration of the potato during the fall and discuss the possibility of the potato breaking upon impact. They also consider the force needed to squeeze the potato and the pressure that their weight exerts on the potato. Ultimately, they decide to test the potato by dropping it and measuring the results.
  • #1
nibbel11
36
2
for my own intressest i thought of a way to look i a potato could survive a 100 metre drop. my
hypothesis was abviously no not it will be potato salad when it hits the ground.
but my calculations turned against me as far as i know. here are they
average potato weight: 0,15 kg
height: 100m

i fought it would be a great start to look at Newtons second law
F=ma
and fill in the acceleration

falldistance=4,9*time2
falldistance is equal to the height because i want to now the change in time for the next formula.
so let's shuffle things around
time=√(100:4,9)=4,5s
so it takes 4,5 seconds to get 100 metre down.
than the fallvelocity=9,8*t=9,8*4,5=44,1m/s

acceleration=(44,1/(the instant moment it touches the ground let's make a assumpsion from 0,1second))=
44,1:0,1=441m/s2

than force=0,15(average potato mass)*441= 66,2N

after i found that answer i put the potato on a electrical scale while pushing it so it says 7kg because
Fz=m*g=7*9,8= 68,6N so i assumed it would simulate the same effect as the drop as the nettoforce would be the same. but while pushing on the potato nothing happened did i go somewhere wrong or is it really possible to drop a potato that high with minor harm

nibbel
 
Physics news on Phys.org
  • #2
You guessed at a deceleration time of one tenth of a second. Why not do a sanity check on that figure? You have already calculated how fast the potato will be moving at impact. How long will it take to travel a distance equal to its diameter at that speed? (or at half that speed if you assume uniform deceleration).
 
  • #3
you mean the diametre of the potato, the potato falls vertically what i think is logical way if you look at its shape, it is more streamlined that way.
i would say something like 8cm and that would be traveled in 0,002 seconds, so let's half it as you said that makes 0,001.
than that would change the acceleration into 44100m/s2
multiply that by 0,15 to get the new resulting force and that would be 6615N
allright but how do i test if the potato handle that force without assuming.
 
  • #4
nibbel11 said:
you mean the diametre of the potato, the potato falls vertically what i think is logical way if you look at its shape, it is more streamlined that way.
i would say something like 8cm and that would be traveled in 0,002 seconds, so let's half it as you said that makes 0,001.
than that would change the acceleration into 44100m/s2
multiply that by 0,15 to get the new resulting force and that would be 6615N
allright but how do i test if the potato handle that force without assuming.
Back up a moment and decide what you are trying to accomplish. The 8 cm figure is just a sanity check. We know that the potato cannot travel more than 8 cm while slowing down because that would turn mean that its top would have to pass through its bottom. Any calculation that involves such a scenario does not pass the sanity check. But a calculation that passes the sanity check may not be correct.

How far can you squeeze a potato before it ruptures? If you have an 8cm potato, can you squeeze it down to 7 cm?
 
  • #5
but even if i knew how far i could sqeeze it, how do i know how much force or pressure i need. because of the force is more than that sqeezing point it would rupture. so what is this "sqeezing point". how do i test this.
 
  • #6
nibbel11 said:
but even if i knew how far i could sqeeze it, how do i know how much force or pressure i need. because of the force is more than that sqeezing point it would rupture. so what is this "sqeezing point". how do i test this.
What are you trying to determine? How far you can drop a potato without having it come apart? The test for that is to drop it and see.
 
  • #7
Use a force plate to land on or scales that log data.

A priori you need to use the kinematic equation involving acceleration and stopping distance, the initial speed just before the deceleration will be the final speed of the trajectory.
 
  • #8
jbriggs i was thinking about what you said earlier and i think i got it.
there is a crumplezone of 1cm
and with a 44.1m/s velocity. and it's average velocity until it reaches 0 will be 22.1m/s
so 0,01:22.1=0,0005s
441:0,0005=882000m/s2
882000*0,15=123300N.
so the force of the fall will be equal to 123300
jbriggs444 said:
The test for that is to drop it and see.
even if i did i still want it on paper.
 
  • #9
nibbel11 said:
and with a 44.1m/s velocity
nibbel11 said:
441:0,0005=882000m/s2
Be careful. It looks like you dropped a factor of ten there.
nibbel11 said:
so the force of the fall will be equal to 123300
That' 123300 Newtons -- with the factor of ten error.

That's on the assumption that the potato decelerates over a distance of one cm. You have not calculated the 1 cm as the actual stopping distance. You have only applied it as an upper bound -- the potato cannot decelerate over any larger distance without breaking. If the acceleration required to stop within that 1 cm is high enough to break the potato then you can conclude that the potato breaks.

So how much is 12300 Newtons? How does it compare to your weight, for instance? Can you stand on a potato without breaking it?
 
  • #10
allright i recalculated the force with the right numbers making a new force of 13230N instead of 123300.
the force on the potato after slowing down is only able to become less than 1323N and compare that to me.
that would be 1323kg on Earth (if it where a mass) and I'm 60kg or 600N, about 20 times less.
i stood on a potato with one foot on the potato's longest side. which had after some calculation 20cm2 contact with my foot.
i did it on it's longest side so i could stand on it and on that position it's area would be the most which gave me the minimum pressure.
so p=F:A=600/20=30N/cm2

the potato will most likely fall on the shortest pointy side if you don't give it to much of an angular momentum.(because of it being more streamlined.
so with that in mind the area wil be about 6cm2
so the pressure the potato tryed to handle at the moment of stopping is 1323/6= 225,5 N/cm2

conlusion:
the potato on which is falling will be likely to become a nonliked christmasmeal because i cannot stand on it and the falling poata's pressure is way more than i did on my potato.
 

1. Will the potato be damaged from the drop?

It is likely that the potato will experience some damage from the drop. The impact of the fall can cause bruising or even breakage of the potato's skin, depending on the height of the drop and the surface it lands on.

2. Is there a specific height that the potato can be dropped without getting damaged?

The height at which a potato can be dropped without getting damaged varies based on factors such as the type of potato, the surface it is dropped on, and the angle of impact. Generally, the higher the drop, the more likely the potato will experience damage.

3. Can the potato be protected from damage during the drop?

It is possible to protect the potato from damage during the drop by using materials such as padding or a container to cushion the impact. However, this may affect the accuracy and reliability of the experiment's results.

4. How will the potato be checked for damage after the drop?

The potato can be visually inspected for any visible signs of damage, such as bruising or breaks in the skin. It can also be weighed before and after the drop to see if there has been any weight loss due to damage to the potato.

5. Why is it important to check if the potato is unharmed after the drop?

The experiment of dropping a potato from a height is often used to demonstrate the impact of gravity on objects. Checking if the potato is unharmed after the drop can help us understand how different forces can affect objects and their structural integrity.

Back
Top