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

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1. Jul 8, 2016

### nibbel11

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 wanna 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

2. Jul 8, 2016

### jbriggs444

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. Jul 8, 2016

### nibbel11

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 travelled 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. Jul 8, 2016

### jbriggs444

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. Jul 8, 2016

### nibbel11

but even if i knew how far i could sqeeze it, how do i know howmuch 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. Jul 8, 2016

### jbriggs444

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. Jul 8, 2016

### houlahound

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. Jul 9, 2016

### nibbel11

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
even if i did i still want it on paper.

9. Jul 9, 2016

### jbriggs444

Be careful. It looks like you dropped a factor of ten there.
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. Jul 9, 2016

### nibbel11

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.