# UP.2.33 The jumping flea problem

• MHB
• karush
In summary: For part b), we could write:h(t)=\frac{t}{2}\left(2v_0-gt\right)where $gt$ is the time taken for the flea to reach its maximum height.
karush
Gold Member
MHB
a. If a flea can jumps straight up to a height of 0.440m.
what is the initial speed as it leaves the ground.
b. For how much time is it in the air?

ok I am sure we use this equation

$$x=x_0+v_o^t+\frac{1}{2} at^2$$

I watched another YT on this but it got to much complication
better just to come here

Apart from YouTube, do you also have a textbook? What does that say?
(I am asking because I am rarely able to understand something from a video that I did not understand previously already.)

I think you mean $v_0t$ in the equation for $x$?

In a drawing, let the vertical axis be the $x$ axis, with upward = positive. The initial $x$-position $x_0 = 0$.
What do you choose for the acceleration $a$? What is the sign of $a$?

Now note that at the maximal $x$-position the velocity equals zero.
How do you obtain an expression for the velocity from the expression you have given?

I got that from the YT vid

the text is University Physics
but equation the is also form 2-12 on their eq list

I not in a class just doing this on my own.

Last edited:
(a) $v_f^2 = v_0^2 -2g \Delta y$

At the top of its trajectory, the flea’s velocity is zero and $\Delta y = h_{max}$ ...

$v_0 = \sqrt{2g h_{max}}$

(b) calculate the time it takes for the flea to reach the top of its trajectory, $h_{max}$, and double it ...

$v_f = v_0 - gt \implies t = \dfrac{v_0}{g}$

total time, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$

skeeter said:
(a) $v_f^2 = v_0^2 -2g \Delta y$

At the top of its trajectory, the flea’s velocity is zero and $\Delta y = h_{max}$ ...

$v_0 = \sqrt{2g h_{max}}$

(b) calculate the time it takes for the flea to reach the top of its trajectory, $h_{max}$, and double it ...

$v_f = v_0 - gt \implies t = \dfrac{v_0}{g}$

total time, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$

ok the textbook answers to this is

a. 2.94 m/s
b. 0.599 s

I'll see if i can plug into get em
$\displaystyle 0.599s=\frac{2\sqrt{2gh_{max}}}{g}$
if $g=-9.8$ and $h_{max}=.44$

Last edited:
karush said:
a. If a flea can jumps straight up to a height of 0.440m.
what is the initial speed as it leaves the ground.
b. For how much time is it in the air?

ok I am sure we use this equation

$$x=x_0+v_ot+\frac{1}{2} at^2$$

I watched another YT on this but it got to much complication
better just to come here
Well, since the "up to a height of 0.44m" is the height above the starting height, you can take $x_0= 0$. Then $x= v_0t+ \frac{1/2}at^2= \frac{1}{2}a(t^2+ \frac{2v_0}{a}t)$ is a quadratic and we can "complete the square": $x= \frac{a}{2}\left(t^2+ \frac{2v_0}{a}t+ \frac{v_0^2}{a^2}- \frac{v_0^2}{a^2}\right)= \frac{a}{2}\left(x+ \frac{v_0}{a}\right)^2- \frac{v_0^2}{2a^2}$. "a" here is negative so that this is an "upside down" parabola. When $t= -\frac{v_0}{a}$ it takes its maximum value $-\frac{v_0^2}{2a}$.

For (a), assuming this is on the surface of the earth, set a= -9.81 and solve $\frac{v_0^2}{19.62}= 0.440$ for $v_0$.

For (b), since a parabola is symmetric about its midline, take t equal to twice the "t" above, $-\frac{2v_0}{a}$.

Last edited by a moderator:
HallsofIvy said:
Well, since the "up to a height of 0.44m" is the height above the starting height, you can take $x_0= 0$. Then $x= v_0t+ \frac{1/2}at^2= \frac{1}{2}a(t^2+ \frac{2v_0}{a}t)$ is a quadratic and we can "complete the square": $x= \frac{a}{2}\left(t^2+ \frac{2v_0}{a}t+ \frac{v_0^2}{a^2}- \frac{v_0^2}{a^2}\right)= \frac{a}{2}\left(x+ \frac{v_0}{a}\right)^2- \frac{v_0^2}{2a^2}$. "a" here is negative so that this is an "upside down" parabola. When $t= -\frac{v_0}{a}$ it takes its maximum value $-\frac{v_0^2}{2a}$.

For (a), assuming this is on the surface of the earth, set a= -9.81 and solve $\frac{v_0^2}{19.62}= 0.440$ for $v_0$.

For (b), since a parabola is symmetric about its midline, take t equal to twice the "t" above, $-\frac{2v_0}{a}$.

2(2.94)/9.81=.599

karush said:
a. If a flea can jumps straight up to a height of 0.440m.
what is the initial speed as it leaves the ground.
b. For how much time is it in the air?

ok I am sure we use this equation

$$x=x_0+v_o^t+\frac{1}{2} at^2$$

I watched another YT on this but it got to much complication
better just to come here

For part a), we could use conservation of energy...initially the flea has kinetic energy, and then when it reaches the apex of its jump, it has potential energy:

$$\displaystyle \frac{1}{2}mv_0^2=mgh$$

Solving for $v_0$, we obtain:

$$\displaystyle v_0=\sqrt{2gh}$$

For part b), we could write:

$$\displaystyle h(t)=\frac{t}{2}\left(2v_0-gt\right)$$

We are interested in the non-zero root, hence:

$$\displaystyle t=\frac{2v_0}{g}=\frac{2\sqrt{2gh}}{g}=2\sqrt{\frac{2h}{g}}$$

## 1. What is the "jumping flea problem"?

The jumping flea problem, also known as the "frog problem" or "grasshopper problem", is a classic mathematical puzzle that involves a flea jumping on a number line. The flea starts at 0 and can only jump in positive integer increments. The goal is to find the minimum number of jumps needed to reach a specific target number, such as 33.

## 2. Why is the jumping flea problem important?

The jumping flea problem is important because it teaches valuable problem-solving skills and mathematical concepts such as number patterns, sequences, and counting. It also helps develop critical thinking and logical reasoning skills.

## 3. How do you solve the jumping flea problem?

The jumping flea problem can be solved using a variety of approaches, such as trial and error, using a number line or a table, or using algebraic equations. One possible solution is to start with the target number and work backwards, subtracting the largest possible jump until you reach 0. The number of jumps needed will be equal to the number of subtractions made.

## 4. Can the jumping flea problem be solved using a formula?

Yes, the jumping flea problem can be solved using a formula. The formula is n(n+1)/2, where n is the target number. For example, if the target number is 33, the formula would be (33)(33+1)/2 = 561/2 = 33, meaning the flea would need 33 jumps to reach 33. However, this formula only works for targets that are triangular numbers (numbers that can be represented as a triangle of dots).

## 5. What are some real-life applications of the jumping flea problem?

The jumping flea problem has real-life applications in various fields such as computer science, physics, and finance. In computer science, it can help with optimizing algorithms and finding the most efficient route for data transmission. In physics, it can be used to model the motion of particles. In finance, it can be used to calculate compound interest or analyze stock market trends.

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