Kinematics motion - Dancers pausing at top of their leap

[negation]

Kinematics motion -- Dancers pausing at top of their leap

Homework Statement

Ice skaters, ballet dancers, and basket ball players executing vertical leaps often give the illusion of "hanging" almost motionless near the top of the leap. To see why this is, consider a leap to maximum height, h. Of the total time spent in the air, what fraction is spent in the upper half (i.e., at y > 0.5h)

Homework Equations

None

The Attempt at a Solution

yf = yi + vit + 0.5gt^2

yf = 0 + vit + 0.5(-9.8ms^-2)t^2
yf = vit - 4.9ms^-2 t^2

Should I find the first derivative of yf and determine the local max? And what does upper half implies? The max time take for the object to travel from ground to max height?

 

[TSny]

Edited: If h is the maximum height of the jump, you want to find the ratio of the time to travel the upper half (from y = .5h to the max height y = h and back to y = .5h) to the total time to travel from the ground and back to the ground.

 

[negation]

By finding the quadratic roots of each of the equation, and taking the time taken for the object to travel up from ground to max height,h and from 0.5h to ma height, h, its possible to find the ratio.
However, I'm unable to reduce the quadratic equations to their respective roots without any numerical values for vi in place.

 

[nasu]

You did not use the fact that h is the maximum height reached when the object is launched with vi.
You need to express the time as a function of h (and g) only. So eliminate vi first.

And of course, in the second equation the initial speed is different. Using the same "vi" for both is confusing.

 

[negation]

you did not use the fact that h is the maximum height reached when the object is launched with vi.
You need to express the time as a function of h (and g) only. So eliminate vi first.

And of course, in the second equation the initial speed is different. Using the same "vi" for both is confusing.

https://www.physicsforums.com/attachments/64943

Is this right?

 

[TSny]

In going from the 2nd to the 3rd line, did you multiply 2(-9.81) correctly?

Don't forget the equation v_f = v_i + at

 

[negation]

In going from the 2nd to the 3rd line, did you multiply 2(-9.81) correctly?

Don't forget the equation v_f = v_i + at

Corrected the blunder.

Is it not possible to solve it via quadratic?

 

[TSny]

For this problem, there is no need to plug in values for g.

You have 0² -v_i² = -2gh.

What do you get if you solve this for v_i in terms of g and h?

-----

Yes, you can work with the quadratic equation in t, but it is easier to work with the linear equation.

 

[negation]

For this problem, there is no need to plug in values for g.

You have 0² -v_i² = -2gh.

What do you get if you solve this for v_i in terms of g and h?

-----

Yes, you can work with the quadratic equation in t, but it is easier to work with the linear equation.

Why is the rational for not plugging in values for g?

v_i= sqrt[-2gh]

 

[haruspex]

View attachment 64944
Corrected the blunder.

Is it not possible to solve it via quadratic?

You're overcomplicating it. We don't need to consider the whole leap. The descent half will do. The ascent is just a mirror image of that in time.
You have the distance, the initial speed (0) and the acceleration. h = gt²/2.
How long does it take to descend half way?

Btw, dancers might intentionally exaggerate the effect by the way they move their limbs. The downside is that the maximum height their heads reach is less. A footballer is unlikely to do that.

 

[TSny]

The expressions are "cleaner" if left in symbols. Leaving the expressions in terms of g will also show that the final answer for this problem does not depend on the value of g. You would get the same answer for someone jumping on the moon as here on the earth.

g is usually defined as the magnitude of the acceleration of gravity, which makes g a positive number.
Note that your result v_i = Sqrt[-2gh] has the square root of a negative number if g is considered positive. [EDIT: However, if you wanted to go against convention and take g to be a negative number throughout the problem, then your equation for v_i is correct.]

[Starting with 0² - v_i² = -2gh, you should not end up with a negative sign inside the square root when solving for v_i.]

 

[negation]

You're overcomplicating it. We don't need to consider the whole leap. The descent half will do. The ascent is just a mirror image of that in time.
You have the distance, the initial speed (0) and the acceleration. h = gt²/2.
How long does it take to descend half way?

Btw, dancers might intentionally exaggerate the effect by the way they move their limbs. The downside is that the maximum height their heads reach is less. A footballer is unlikely to do that.

it takes t = sqrt[2h/g]
I find it easier to find the ratio between time taken to travel from ground to h and time taken to travel from h/2 to h.

by the way, could you explain to me how h = gt^2 is derived?

 

[negation]

The expressions are "cleaner" if left in symbols. Leaving the expressions in terms of g will also show that the final answer for this problem does not depend on the value of g. You would get the same answer for someone jumping on the moon as here on the earth.

g is usually defined as the magnitude of the acceleration of gravity, which makes g a positive number.
Note that your result v_i = Sqrt[-2gh] has the square root of a negative number.

Starting with 0² - v_i² = -2gh, you should not end up with the square root of a negative number when solving for v_i.

I understand. But because I defined gravity to be negative if downwards,
v_i = Sqrt[-2(-)h]
v_i will give me a positive value.

 

[TSny]

I understand. But because I defined gravity to be negative if downwards,
v_i = Sqrt[-2(-)h]
v_i will give me a positive value.

OK. That's fine. It's just not the usual way people use the symbol g.

 

[negation]

OK. That's fine. It's just not the usual way people use the symbol g.

I know. I try not to make too much unnecessary arbitrary distinction in the equation. It's less chaotic this way.

Can we continue?

 

[TSny]

it takes t = sqrt[2h/g]

This is correct for the time to go from the ground up to maximum height.

by the way, could you explain to me how h = gt^2 is derived?

This is not a correct equation.

 

[negation]

This is correct for the time to go from the ground up to maximum height.



This is not a correct equation.

Sorry. h = [gt^2]/2
How is it derived?

 

[TSny]

Sorry. h = [gt^2]/2
How is it derived?

That was brought in by haruspex for his approach to the problem. I'll let him/her explain it.

 

[negation]

that was brought in by haruspex for his approach to the problem. I'll let him explain it.

 

[TSny]

Your result $t = -\frac{\sqrt{-2gh}}{g}$ is correct (assuming your convention that $g$ is a negative number).

With the usual convention, you would have $t = \frac{\sqrt{2gh}}{g}$. You can simplify the result by taking the $g$ in the denominator inside the square root and doing some canceling.

 

[negation]

Your result $t = -\frac{\sqrt{-2gh}}{g}$ is correct (assuming your convention that $g$ is a negative number).

With the usual convention, you would have $t = \frac{\sqrt{2gh}}{g}$. You can simplify the result by taking the $g$ in the denominator inside the square root and doing some canceling.

sure.
Now that we have the time taken for object to travel from 0m to height, h.

In determining time taken for object to travel from h/2 to height, h.

 

[TSny]

Look at your first equation: $h -h/2 = .5(\sqrt{-2gh} + 0)*t$. It appears that you are saying that the displacement equals the average velocity multiplied by the time. That's one way to get it. But what does $\sqrt{-2gh}$ in this equation represent?

 

[negation]

Look at your first equation: $h -h/2 = .5(\sqrt{-2gh} + 0)*t$. It appears that you are saying that the displacement equals the average velocity multiplied by the time. That's one way to get it. But what does $\sqrt{-2gh}$ in this equation represent?

It represents v_i.

 

[TSny]

Does v_i here represent the velocity at the ground level or the velocity at h/2?

 

[negation]

Does v_i here represent the velocity at the ground level or the velocity at h/2?

v_i represents velocity at position h/2

 

[TSny]

OK, v_i should be the velocity at h/2. But the expression $\sqrt{-2gh}$ is not the velocity at h/2, it's the velocity at the ground.

 

[negation]

ok, v_i should be the velocity at h/2. But the expression $\sqrt{-2gh}$ is not the velocity at h/2, it's the velocity at the ground.

 

[TSny]

OK. Note that the 2's will cancel in your final expression.

 

[negation]

OK. Note that the 2's will cancel in your final expression.

Is there a single equation that allows me to determine the change in v_i as the object travels from 0m to h/2 without having to produce the above equation?

 

[TSny]

Seems like that question is leading us aside. We have a couple of choices. We can continue your line of thinking (which is getting you pretty close to the result) or we can step back and start all over with a different approach that will lead to the answer with the least amount of calculation (for example, haruspex's method).

Which would you like to do?

 

[negation]

Seems like that question is leading us aside. We have a couple of choices. We can continue your line of thinking (which is getting you pretty close to the result) or we can step back and start all over with a different approach that will lead to the answer with the least amount of calculation (for example, haruspex's method).

Which would you like to do?

I'll like to finish up with my line of reasoning.

Then run another loop using a different method.

 

[TSny]

OK, it's getting pretty late here where I am. But let's finish up your line of reasoning.

So, going back to $h-h/2=.5(v_i+v_f)t$, what do you get for t after using your result for $v_i$?

[EDIT. Oops , I see you already have done this. But you made a mistake in the algebra. h should appear in your answer for t.]

 

[negation]

OK, it's getting pretty late here where I am. But let's finish up your line of reasoning.

So, going back to $h-h/2=.5(v_i+v_f)t$, what do you get for t after using your result for $v_i$?

As above

If the above is true, then, what follows is basic division and algebraic manipulation.

 

[TSny]

Your answer for t should contain h as well as g.

 

[negation]

Your answer for t should contain h as well as g.

It did until I did some cancelling