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Physics UP.2.33 The jumping flea problem

karush

Well-known member
Jan 31, 2012
3,004
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
 

Janssens

Well-known member
Sep 16, 2017
204
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?
 

karush

Well-known member
Jan 31, 2012
3,004
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.
 
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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
980
(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, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$
 

karush

Well-known member
Jan 31, 2012
3,004
(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 text book answers to this is

a. 2.94 m/s
b. 0.599 s

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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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, $-\frac{2v_0}{a}$.
 
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karush

Well-known member
Jan 31, 2012
3,004
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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