Motion in straight line in bungee jumping

[Kevin98]

Homework Statement

Dito will do bungee jumping. He will jump from a height of 47 meters above the water surface. When he jumps, he will be thrown back when the rope shortens and will oscillate up and down. Dito's speed when he fell was 70 km/hour with a slope level of almost 180⁰. How long does it take for the dito to touch the surface of the water?

Relevant Equations

vf = vi + at
vf² = vi² + 2as

vf²= vi² +2as

vf² = (19,44)² + 2(9,81)(47)

vf = 36,05 m/s

to find the t, we can use formula:

t = (vf - vi)/a

t = (36,05 - 19,44)/9,81

t = 1,693 s.

Is it correct?

 

[pasmith]

You might want to show the working for the conversion from km/h to m/s.

 

[Kevin98]

You might want to show the working for the conversion from km/h to m/s.

70 km/h = (70.000 m/3600 s) = 19,44 m/s

 

[Steve4Physics]

t = 1,693 s.

Is it correct?

Just a few comments in addition to @pasmith's...

Are you sure the question is correctly stated? How can Dito realistically have an initial velcoity of 70km/h downwards? That's fast.

Check the number of significant figures in your final answer.

It may be worth stating the sign-convention you are using (what direction is positive).

An alternative method is to use ##s =v_i t + \frac 12 at^2## - but that means you have to solve a quadratic equation. If you don't want to do that, you can still use the equation to check that your answer gives the correct distance for the time you have calculated.

 

[Kevin98]

Just a few comments in addition to @pasmith's...

Are you sure the question is correctly stated? How can Dito realistically have an initial velcoity of 70km/h downwards? That's fast.

Check the number of significant figures in your final answer.

It may be worth stating the sign-convention you are using (what direction is positive).

An alternative method is to use ##s =v_i t + \frac 12 at^2## - but that means you have to solve a quadratic equation. If you don't want to do that, you can still use the equation to check that your answer gives the correct distance for the time you have calculate

Just a few comments in addition to @pasmith's...

Are you sure the question is correctly stated? How can Dito realistically have an initial velcoity of 70km/h downwards? That's fast.

Check the number of significant figures in your final answer.

It may be worth stating the sign-convention you are using (what direction is positive).

An alternative method is to use ##s =v_i t + \frac 12 at^2## - but that means you have to solve a quadratic equation. If you don't want to do that, you can still use the equation to check that your answer gives the correct distance for the time you have calculated.

Yep. 70 km/h. I also think that it is too fast, but that is what the question says.

 

[Steve4Physics]

In post #4 I should also have said that you are assuming the acceleration is a constant 9.81m/s² downwards. You have calculated the time taken for an object to free-fall 47m.

But there is a bungee-rope - so the acceleration will not be 9.81m/s² downwards all the time.

It sounds like there is a problem with the question. I'd recommend checking.

 

[Kevin98]

In post #4 I should also have said that you are assuming the acceleration is a constant 9.81m/s² downwards. You have calculated the time taken for an object to free-fall 47m.

But there is a bungee-rope - so the acceleration will not be 9.81m/s² downwards all the time.

It sounds like there is a problem with the question. I'd recommend checking.

Is it possible that what the question means by "Dito's speed was 70 km/h when he fell" is the speed when he hit the water surface and the initial speed was zero? If we go by this assumption, the acceleration is definitely not 9,81 m/s². I think you're right and there is a problem with the question itself. I found this question in a chapter about motion kinematics. So, I assume there won't be analysis about different forces acting on an object. What do you think? I also find another problematic question and solution in this book, which I will post later.

 

[erobz]

Is it possible that what the question means by "Dito's speed was 70 km/h when he fell" is the speed when he hit the water surface and the initial speed was zero? If we go by this assumption, the acceleration is definitely not 9,81 m/s². I think you're right and there is a problem with the question itself. I found this question in a chapter about motion kinematics. So, I assume there won't be analysis about different forces acting on an object. What do you think? I also find another problematic question and solution in this book, which I will post later.

Since this is a kinematics problem, I think the interpretation could be the ## 70 ~\rm{km/hr}## was the jumpers maximum speed (just before the bungee engages). Then perhaps they accelerate to rest in the remaining distance at a constant rate.

 

[Steve4Physics]

Is it possible that what the question means by "Dito's speed was 70 km/h when he fell" is the speed when he hit the water surface

If so, I hope Dito sued whoever organised the jump!

I would guess that the intended question is:
a) initial speed = 0;
b) 70 km/h is the maximum speed reached;
c) the bungee rope then causes deceleration;
d) the speed reaches 0 after a total distance of 47m has been covered.

The problem is then what value to use for deceleration. But, as you say, this is then no longer a kinematics equation. [EDIT - but see Post #10.]

If there is an 'official answer' (even if only a single value) given, it would be interesting to know what it is. It might then be possible to work backwards to find the intended question.

I think you're right and there is a problem with the question itself. I found this question in a chapter about motion kinematics. So, I assume there won't be analysis about different forces acting on an object. What do you think?

I agree. [EDIT - but see Post #10.]

I also find another problematic question and solution in this book, which I will post later.

Yippee.

Edit - minor corrections.

 

[Steve4Physics]

Note that, if the rope obeys Hooke's law, the motion when the rope is in tension is simple harmonic. If you have studied simple harmonic motion (SHM) it gives you a way to proceed without considering forces and mass (assuming 70 km/h is Dito's max speed).

 

[kuruman]

This problem statement is incomplete. We are told that the distance to the water is 47 m. That's fine. We are also told that "When he jumps, he will be thrown back when the rope shortens ##\dots## My questions are (a) At what distance from the water does the rope shorten? (b) How stiff is the rope (spring constant)?

 

[erobz]

Note that, if the rope obeys Hooke's law, the motion when the rope is in tension is simple harmonic. If you have studied simple harmonic motion (SHM) it gives you a way to proceed without considering forces and mass (assuming 70 km/h is Dito's max speed).

If the origin is at the point where the bungee engages, then:

$$ x(t) = \frac{v_o}{\omega} \sin ( \omega t ) $$

Seems like there are more variables than equations for that, but maybe I'm missing something?

 

[Steve4Physics]

If the origin is at the point where the bungee engages, then:

$$ x(t) = \frac{v_o}{\omega} \sin ( \omega t ) $$

Seems like there are more variables than equations for that, but maybe I'm missing something?

I think this is OK...

The speed ##v_{max}## = 70km/h when the bungee engages (say extension ##x=0##).

The speed is zero at the water surface; so the amplitude of SHM is ##A =## [distsance between the point where ##x=0## and water surface]
.
##v_{max} = 2 \pi f A## (a standard equation in SHM)

One quarter of a full cycle is performed between the point where x=0 and the first encounter with the water surface.

Or maybe I've said too much.

EDIT: I certainly have said too much! What I said is wrong - see later Posts.

 

[erobz]

I think this is OK...

The speed ##v_{max}## = 70km/h when the bungee engages (say extension ##x=0##).

The speed is zero at the water surface; so the amplitude of SHM is ##A =## [distsance between the point where ##x=0## and water surface]
.
##v_{max} = 2 \pi f A## (a standard equation in SHM)

One quarter of a full cycle is performed between the point where x=0 and the first encounter with the water surface.

Or maybe I've said too much.

Ok, the other equation is the velocity. Thanks!

 

[Kevin98]

Does it go like this?
Dito's initial speed vi1 = 0
Dito's max speed vi2 = 70 km/h = 19,44 m/s (before the bungee engages, acceleration = 9,81 m/s²)
The initial distance covered = the rope's lenght (h1) :
(vi2)² = (vi1)² + 2gh1
(19,44)² = 0 + 2(9,81)(h1)
h1 = (377,91)/19,62
h1 = 19,26 m
Time to reach a speed of 70 km/h
t1 = vi2/g
t1 = 19,44/9,81
t1 = 1,98 s
The remaining distance (after the bungee engages) :
h2 = 47 - 19,26 = 27,74 m
Deceleration :
a = (vi2)²/2h2
a = (377,91)/2(27,74)
a = 13,62 m/s²
Since his speed when he hit the water surface is equal to zero, then:
0 = vi2 - at
t2 = (19,44)/(13,62)
t2 = 1,42 s
total time = t1 + t2 = 1,98 + 1,42 = 3,4 s.
Is it correct?

 

[erobz]

Does it go like this?
Dito's initial speed vi1 = 0
Dito's max speed vi2 = 70 km/h = 19,44 m/s (before the bungee engages, acceleration = 9,81 m/s²)
The initial distance covered = the rope's lenght (h1) :
(vi2)² = (vi1)² + 2gh1
(19,44)² = 0 + 2(9,81)(h1)
h1 = (377,91)/19,62
h1 = 19,26 m
Time to reach a speed of 70 km/h
t1 = vi2/g
t1 = 19,44/9,81
t1 = 1,98 s
The remaining distance (after the bungee engages) :
h2 = 47 - 19,26 = 27,74 m
Deceleration :
a = (vi2)²/2h2
a = (377,91)/2(27,74)
a = 13,62 m/s²
Since his speed when he hit the water surface is equal to zero, then:
0 = vi2 - at
t2 = (19,44)/(13,62)
t2 = 1,42 s
total time = t1 + t2 = 1,98 + 1,42 = 3,4 s.
Is it correct?

Assuming it is supposed to just be a kinematics problem, it's one plausible interpretation. As has been pointed out the question is poorly posed. Another could be using the equations of simple harmonic motion to find the time of the last portion (as @Steve4Physics points out) if you have studied that?

For future reference, algebraic manipulations are preferred until the last step, and the site has math formatting capabilities. Please see: LaTeX Guide.

 

[haruspex]

Dito's max speed vi2 = 70 km/h = 19,44 m/s (before the bungee engages, acceleration = 9,81 m/s²)
The initial distance covered = the rope's lenght (h1) :
(vi2)² = (vi1)² + 2gh1
(19,44)² = 0 + 2(9,81)(h1)

Why do you think the max speed is when the bungee first becomes taut? Will the acceleration instantly change from g to zero?

 

[erobz]

Why do you think the max speed is when the bungee first becomes taut? Will the acceleration instantly change from g to zero?

Do we have an equation to invoke for rate of change of acceleration over the transition?

 

[haruspex]

Do we have an equation to invoke for rate of change of acceleration over the transition?

SHM.

 

[erobz]

SHM.

While that may be consistent with the SHM interpretation ( which may be the intention ), that is not consistent with the model the OP chose to evaluate (the one you quote).

However, if we invoke SHM, then at the onset of the spring force ##o## the velocity of the jumper is then:

$$v(t) = v_o \cos ( \omega t ) $$

Which appears to be maximized at ##t = 0##, with ##v(0) = v_o ##. What am I missing?

 

[Steve4Physics]

While that may be consistent with the SHM interpretation ( which may be the intention ), that is not consistent with the model the OP chose to evaluate (the one you quote).

However, if we invoke SHM, then at the onset of the spring force ##o## the velocity of the jumper is then:

$$v(t) = v_o \cos ( \omega t ) $$

Which appears to be maximized at ##t = 0##, with ##v(0) = v_o ##. What am I missing?

See my Post #13!

 

[erobz]

See my Post #13!

About what though? @haruspex seems to be implying in #19 the velocity of the jumper increases after the spring begins to stretch beyond that of their initial freefall velocity to that point? Maybe I have the wrong equation of motion.

 

[haruspex]

See my Post #13!

With which I disagree. See post #17.
Max velocity is when the net force ….

 

[erobz]

With which I disagree. See post #17.
Max velocity is when the net force ….

Yeah, I have the wrong equation of motion. The velocity is maximized when the acceleration is ##0##. I saw some pictures on wiki, but I didn't digest if the equation I grabbed for ##x(t)## were applicable here. I solved the ODE now, I see the blunder. Shameful approach!

 

[Steve4Physics]

With which I disagree. See post #17.
Max velocity is when the net force ….

Aha. Max velocity occurs when the tension first equals the weight. Up until that point, the net force (and hence acceleration) is downwards. So the speed increases until that point.

 

[erobz]

So... based on the actual equation of motion, solving system as I was trying to before using position and velocity at the water sure doesn't seem like it's going to be analytical?

Given the challenges surrounding the "SHM solution", I suspect the intended solution is as the OP has shown in #15.

 

[erobz]

I did find an analytical solution to the SHM problem after all. It's significantly more work for about ##0.4 ~\rm{s}## increase on total time when compared to the "constant force" approach in #15. I guess it's not insignificant...but it's pretty close for this one.