
MHB Master
#1
December 19th, 2017,
20:27
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

December 19th, 2017 20:27
# ADS
Circuit advertisement

MHB Craftsman
#2
December 19th, 2017,
20:44
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?

MHB Master
#3
December 19th, 2017,
21:29
Thread Author
I got that from the YT vid
the text is University Physics
but equation the is also form 212 on their eq list
I not in a class just doing this on my own.
Last edited by karush; December 19th, 2017 at 23:15.

MHB Craftsman
#4
December 20th, 2017,
09:07
(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, $T=\dfrac{2v_0}{g} = \dfrac{2\sqrt{2gh_{max}}}{g}$

MHB Master
#5
December 20th, 2017,
16:35
Thread Author
â€‹
Originally Posted by
skeeter
(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 $h_{max}=.44$
Last edited by karush; December 20th, 2017 at 17:14.

#6
December 21st, 2017,
21:48
Originally Posted by
karush
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 HallsofIvy; December 23rd, 2017 at 16:08.

MHB Master
#7
December 28th, 2017,
16:54
Thread Author
Originally Posted by
HallsofIvy
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

Pessimist Singularitarian
#8
December 29th, 2017,
07:49
Originally Posted by
karush
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_0gt\right)$
We are interested in the nonzero root, hence:
$ \displaystyle t=\frac{2v_0}{g}=\frac{2\sqrt{2gh}}{g}=2\sqrt{\frac{2h}{g}}$