# Parabolic motion problem

1. May 2, 2012

### Perpendicular

1. The problem statement, all variables and given/known data

a hyper-intelligent flea jumps over a trunk with radius R lying on the ground assuming the shape of a cylinder. What minimum velocity must it exert to land on the other side , clearing the trunk ?

2. Relevant equations

displacement along x-axis = v2/g sin2x where x = angle of projection.

3. The attempt at a solution

I assume that the circular cross section of the trunk cuts the parabolic trajectory at its highest point, but this gives me an equation for the y-component of the vector. How do I get the x-component equations ?

2. May 2, 2012

### tiny-tim

Hi Perpendicular!

Hint: what is the largest circle that will fit at the bottom of a parabola?

3. May 2, 2012

### andrien

so that in that much time it just clears of a horizontal distance of 2R

4. May 2, 2012

### Perpendicular

the end and start points of the parabola have a distance 2R ? but the circle is in the middle of the parabola, isn't it ? that means the horizontal diametric distance along the circle can at best approach the parabola, and equal it when the whole diagram looks like a pulley system. but how is that the case ?

5. May 2, 2012

### Perpendicular

not sure what you mean. I've placed the circular tree trunk in the middle to maximize that, and as the flea needs to just jump over I'm assuming the parabola's peak point is at the circle's highest point too.

6. May 2, 2012

### tiny-tim

but how wide does the parabola have to be?

7. May 2, 2012

### andrien

can you clarify what data is given,means if the angle of projection or something else.in what terms do you want v

8. May 2, 2012

### Perpendicular

v should be in terms of g and R.

>2R.

9. May 3, 2012

### andrien

v=√(17/4gR) , tanθ=4

10. May 3, 2012

### Perpendicular

Answer given as 2.197 x Sqrt(gR). Not exactly root of 17/4...what was your method, though ?

11. May 8, 2012

### andrien

that answer is wrong.sorry,but at my best i can find 2.236√gR WHICH IS IN FACT √5gR which I have gotten using the vertical eqn you are using.and for horizontal I have fitted a parabola
which passes through the topmost point and for any given vertical distance the difference in value of horizontal (x) for parabola and the circle is minimum parabola lying always outside the circle.the answer seems to imply that your vertical eqn does not hold perhaps in some way getting a compromise in horizontal and vertical will give a minimum velocity which is somewhat smaller than I am getting when I am using your answer the parabola goes inside of cylinder.moreover can you give me reference for the problem and i hope tim will have some way out of it.

12. May 8, 2012

### tiny-tim

say you have the parabola y = x2

obviously a tiny circle will easily fit in the bottom (at the origin), but a large circle won't …

so what is the largest circle that will fit?

13. May 8, 2012

### Perpendicular

the problem is from a NSEP paper...national standard exam in physics in India.

anyway nevermind, turns out there is a solution online :

http://scitation.aip.org/journals/doc/PHTEAH-home/challenges/nov2010.pdf [Broken]

Last edited by a moderator: May 6, 2017
14. May 8, 2012

### tiny-tim

let's hope you have open-laptop exams!

15. May 8, 2012

### andrien

all right,I was thinking that through as I have written there that there will be a compromise between vertical and horizontal motion.your vertical direction assumption was wrong.nevertheless that was a good one.