Question

In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/s2 . Ignore air resistance.

A flea jumps straight up to a maximum height of 0.460m . What is its initial velocity v0 as it leaves the ground?

Answer 1

\[\frac{ mv^2 }{ 2 }=mgh \rightarrow v=\sqrt{2gh}=3m/s\]

Answer 2

yeah, there is no way i understand that... you didn't explain @Fifciol

Answer 3

Well, I don't understand how you derived that equation

Answer 4

okay, at the top of the jump, all of the energy is potential energy, right? the flea is briefly motionless before it starts going back down.

Answer 5

It's the conservation of mechanical energy. Assuming that there is no air resistance the energy must be conserved. The initial velocity is then the same as the velocity of an object when it hits the ground.

Answer 6

that means the potential energy is \(mgh\) where \(m\) is mass, \(g\) is gravitational acceleration, \(h\) is height.

Answer 7

can i post something hold on

Answer 8

the left side of the equation is kinetic energy. the kinetic energy when the flea jumps is all turned into potential energy at the top of the jump (because we are assuming no air resistance, etc). there is no potential energy when the flea jumps, but as it rises, the kinetic energy turns into potential energy until it is all potential energy.

Answer 9

you'll do lots of physics problems with a similar setup: total energy remains constant, it just shifts between potential and kinetic.

Answer 10

these are the only equations I am allowed to use, I dont understand how you are getting the other equation

Answer 11

\[W=\int\limits_{a}^{b}Fdx=\int\limits_{a}^{b}m \frac{ dv }{ dt }dx=\int\limits_{v_1}^{v_2}mvdv=\frac{ m \Delta v^2 }{2 } \]

Answer 12

im sorry, that doesn't make any sense to me

Answer 13

You can solve this problem using simple 1D kinematics:
\[v=v_0 -gt\] the minius sign indicates that the acceleration is in the opposite sign with the increasing values of v. in this case v=0 so you get \[v_0=gt\]
\[h=\frac{ g t^2 }{ 2 }\rightarrow t=\sqrt{\frac{ 2h }{ g }}\]
substitute that in equation of v zero and you have the same answer

Answer 14

where did you get the 2?

Answer 15

\[x = \frac{1}{2}at^2 = \frac{1}{2}gt^2\]

Answer 16

im still lost

Answer 17

it's the final term in your equation
\[y = y_0+v_0 t -\frac{1}{2}gt^2\]

Answer 18

\(y_0\) is initial position
\(v_0\) is initial velocity

Answer 19

so if the flea starts falling from a height of 0.460m, with no initial speed, the equation says that its height is
\[y = 0.460\text{ m} + 0t - \frac{1}{2}{(9.8 \text{ m/s}^2 )} t^2\]

Answer 20

Here's a graph of the flea's height (y-axis) as a function of time (x-axis)

Answer 21

but there are 2 unknowns, how do we solve that

Answer 22

you can see it slowly picks up speed and is moving much faster as it hits the ground (in fact, it buries itself in the ground, it appears :-)

Answer 23

so we could work out the time it takes to fall by finding the value of \(t\) where \(y = 0\). however, we don't need to do that for this problem.

Answer 24

because there is no air resistance, the trajectory of the flea is symmetrical — it hits at the same speed as it jumps...

Answer 25

@summersnow8 you said there are 2 unknowns, what do you think they are?

Answer 26

well, we dont know y and we dont know t, how do we solve the problem....

Answer 27

y is a function of t. at any given value of t, we can find y by evaluating
\[y = y_0 + v_0 t - \frac{1}{2}gt^2\]where \(y_0\) is the initial height from which freefall starts, and \(v_0\) is the initial velocity (which is 0).

Answer 28

use third equation
\[v^2=v_0^2-2g(y-y_0)\] where \[y-y_0=h\]
v^2 =0 so again you get the answer

Answer 29

sometimes you see that third equation written as
\[v_f^2 - v_i^2 = 2 a d\] where \(a\) is acceleration and \(d\) is distance travelled between the spot where we were going \(v_i\) and the spot where we were going \(v_f\)

Answer 30

but we still have \[y= 0.460 - \frac{ 1 }{ 2 }\left( 9.8 \right)\t^2]

Answer 31

yes. and at t = 0 of the freefall, that simplifies to 0.460 m, the height at which the flea is at the top of the jump.

Answer 32

how do we know t=0, it doesnt say that t=0

Answer 33

because the freefall starts at t=0. otherwise, we'd already be moving and in a different spot. any value of y after t = 0 is below the starting height if v_0 is 0 (as it is here) or negative (as it would be if the flea was ejected downward from something).

Answer 34

so now we have y=.460m

Answer 35

there are three components to that equation.

the first component is the y_0 component — that's where we start.
the second component is the component that gives our displacement due to any initial velocity. if this is present and there's no net acceleration, then the position just goes steadily up or down, with the slope being the value of v_0

the third component is the component due to the acceleration (by gravity in this case). it gets larger and larger as time goes on, until the object finally comes to rest

Answer 36

if you just have the first component, the object is sitting there, suspended in space, not moving. no matter the value of t, the position is the same.

if you add the second component, we get constant motion.

add in the third component and the motion changes with time

Answer 37

im not following, and there is B and C..... :/

Answer 38

okay. let's look at those formulas.
\[v^2=v_0^2-2g(y-y_0)\] this says that the velocity can be found if you know the initial velocity, the acceleration due to gravity, starting position, and ending position. we know all of those things, right?

Answer 39

i don't know anymore

Answer 40

when the flea jumps, he jumps at speed \(v_0\). at the top of the jump, he is momentarily at speed \(v = 0\). height goes from \(y_0 = 0\) to \(y = 0.460\)

Answer 41

if we plug in the numbers,

\[(0)^2 = v_0^2 - 2(9.8)(0.460-0)\]\[v_0^2 = 2(9.8)(0.460)\]Take square root of both sides
\[v_0 = \sqrt{2(9.8)(0.460)} = \sqrt{9.016} = 3.0 \text{ m/s}\] (to 2 sig figs)

Answer 42

sorry, I see that's 9.80 in your problem statements, so it should be 3.00 m/s (3 sig figs)

Answer 43

i guess

Answer 44

well, let's back up if you're not comfortable. where do you stop feeling comfortable?

Answer 45

identifying, and applying

Answer 46

identifying the values of variables, or the equation(s) to use?

Answer 47

You highlighted 4 equations:
Equation #1
\[v = v_0 - gt\]Equation #2\[y = y_0 + v_0 t - \frac{1}{2}gt^2\]Equation #3\[v^2 = v_0^2 - 2 g(y-y_0)\]Equation #4\[y = y_0 + \frac{1}{2}(v+v_0)t\]

Equation #3 was the one to use here because it only involves quantities that we already knew \((v,g,y,y_0)\), and of course the one we wanted to find \((v_0)\).

Answer 48

i give up

Answer 49

We can use Eq. 1 to find the velocity of an object in freefall if we know the elapsed time and the initial velocity. We didn't know the elapsed time, so it wouldn't be the easiest route.

We can use Eq. 2 to find the position of an object in freefall if we know initial position, initial velocity, and elapsed time. Unfortunately, we don't know elapsed time.

We can use Eq. 4 to find the position of an object in freefall if we know initial position, velocity, initial velocity, and elapsed time. We still don't know the the elapsed time.

Any of these equations could be rearranged and used to find a different member of the set of variables they relate, of course. In Eq. 1, we could find the initial velocity if we knew the velocity and the elapsed time, for example.

Answer 50

Instead of giving up, why don't we look at the rest of the problem?

Answer 51

didn't you say there were some other parts?

Answer 52

I dont understand physics

Answer 53

not even physicists do :-)

Answer 54

the point of taking the course is to improve your understanding, not to demonstrate that you already know the material...

Answer 55

it's like riding a bicycle up a hill. if it is easy, you just aren't going fast enough :-)

Answer 56

I can't do this right now

Answer 57

okay. you know where to find me...

Answer 58

there is no way I will understand, no matter how many times you explain it to me

Answer 59

If you cling to that sentiment, it will become a self-fulfilling prophecy. Personally, I prefer to think of things that seem incomprehensible today as things that might make sense tomorrow, if I take a fresh look. Who wants to go through life thinking "I can't do that"?

As Henry Ford so eloquently put it, "if you think you can, or you think you can't, you're probably right." http://www.psychologytoday.com/blog/flourish/201002/if-you-think-you-can-t-think-again-the-sway-self-efficacy

Answer 60

a=9.80,s=0.460,v=0,find u
v^2=U^2+2as
0=u^2+2(9.80)(0.460)
u=3.002665