Free fall question, where has my reasoning gone wrong?

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The discussion centers around a physics problem involving a bolt dropped from an elevator that is moving upward. The initial confusion arises from the calculation of the time it takes for the bolt to hit the ground, with participants arriving at different answers due to the effects of the elevator's upward acceleration and the bolt's initial velocity. It is clarified that the bolt retains the elevator's upward velocity of 8 ft/s when dropped, which must be considered in the calculations. The discrepancy in the time of flight answers (0.71 seconds from the book versus 0.75 seconds from calculations) is attributed to the need to account for the elevator's motion, not just gravity. The conversation emphasizes understanding the relationship between the bolt's initial velocity and the forces acting on it after being released.
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Homework Statement


It's a two part question, but it basically says this:

a bolt drops from an elevator 9.0 ft from the ground, at what time does it hit the ground?


Homework Equations



x = x0 + vx0t + 1/2ax(t^2)


The Attempt at a Solution



since it's a free fall question and we're starting from above ground:

0 = 9.0ft - 16ft/s^2 (t^2)

-9.0/-16 = t^2

t = 0.75 seconds.

But the book says that t = 0.71 seconds..

thanks
 
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DaveC426913 said:
Where does the 16ft/s^2 come from?


gravity is 32 ft/s^2 so 1/2ax(t^2) is -16ft/s^2(t^2)

thanks for replying
 
Your answer looks fine to me.
 
a bolt drops from an elevator 9.0 ft from the ground, at what time does it hit the ground?

so...

1 foot = 0.3048 m

so 9ft = 2.7432 m

(delta) y = Vot + 1/2(a)t^2

so 2.7432 = 0 + 1/2(-9.80)t^2

so 2.7432/(1/2*-9.8)=t^2

so .5598367347 = t^2

so t = .7482223832 or .75s
 
Doc Al said:
Your answer looks fine to me.

how come the answer says 0.71 seconds though? the equation sqrt(9/16) is not ambiguous, since they are both perfect squares.. I can't really guess that it was a difference due to rounding can I?

thanks
 
emyt said:
how come the answer says 0.71 seconds though? the equation sqrt(-9/-16) is not ambiguous, since they are both perfect squares.. I can't really guess that it was a difference due to rounding can I?

thanks

my math got me .75s (I stored the answers in my calculator so rounding error wouldn't be a factor)

the book probably rounded each answer or rounded off wrong... your answer is correct
 
emyt said:
gravity is 32 ft/s^2 so 1/2ax(t^2) is -16ft/s^2(t^2)

thanks for replying
Yeah. I realized as soon as I posted, so I deleted it so as not to confuse.
 
bleedblue1234 said:
my math got me .75s (I stored the answers in my calculator so rounding error wouldn't be a factor)

that's what I got as well, you don't need a calculator to see that the sqrt( 9 / 16) is going to be 0.75.. but the problem is that the book says 0.71s, which is strange...
 
What is the other part of this "two part question"? Is it possible that the 9.0 ft initial height was derived and thus possibly incorrect?
 
  • #10
DaveC426913 said:
What is the other part of this two part question? Perhaps that factors in?

this is the full question:

An elevator ascends with an upward acceleration of 4.0ft/s^2. At the instant its upward speed is 8.0ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor.

Calculate (a) the time of flight of the bolt from the ceiling to the floor

(b) the distance it has fallen relative to the elevator shaft.

Answer: (a) 0.71 (b) 2.3 ft

I don't see how the other parts can affect the answer.. but there's a large chance that I'm wrong, I'm just learning physics myself through this book

thanks
DaveC426913 said:
What is the other part of this "two part question"? Is it possible that the 9.0 ft initial height was derived and thus possibly incorrect?

no, it is given
 
  • #11
emyt said:
this is the full question:

An elevator ascends with an upward acceleration of 4.0ft/s^2. At the instant its upward speed is 8.0ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor.

Calculate (a) the time of flight of the bolt from the ceiling to the floor
That's quite a bit different from the problem that you stated and solved! You can't ignore the acceleration of the elevator.
 
  • #12
Doc Al said:
That's quite a bit different from the problem that you stated and solved! You can't ignore the acceleration of the elevator.

hmm, but that is the acceleration going up isn't it? when it falls, wouldn't the acceleration just be gravity?

thanks
 
  • #13
emyt said:
hmm, but that is the acceleration going up isn't it? when it falls, wouldn't the acceleration just be gravity?

thanks

Yes but it's initial velocity is not zero. And the elevator floor is moving.


This is not a two-part question. This is a one-part question that has several dependent components.
 
  • #14
emyt said:
hmm, but that is the acceleration going up isn't it? when it falls, wouldn't the acceleration just be gravity?
The acceleration of the bolt is just due to gravity. But you are asked for the time until it hits the moving floor, which is accelerating. You need to combine both accelerations. (Write the position of the bolt and the elevator--with respect to the ground--as functions of time. Then solve for the point at which they collide.)
 
  • #15
ohh. I thought that they meant the floor of the elevator SHAFT. thanks.

I thought that a dropped object has an initial velocity of 0? If not, I'm not sure how I would calculate that.

Thanks for the replies
 
  • #16
emyt said:
ohh. I thought that they meant the floor of the elevator SHAFT. thanks.
D'oh! :wink:

I thought that a dropped object has an initial velocity of 0? If not, I'm not sure how I would calculate that.
The dropped object will have the initial speed of whatever it's being dropped from. (In this case from a moving elevator.)
 
  • #17
Doc Al said:
D'oh! :wink:


The dropped object will have the initial speed of whatever it's being dropped from. (In this case from a moving elevator.)

A good way to think about this would be to think about what happens in a car. When you drop something out of the window, it keeps moving with the car's velocity, it doesn't start from rest. ;)
 
  • #18
emyt said:
I thought that a dropped object has an initial velocity of 0? If not, I'm not sure how I would calculate that.
Well your equation contains a variable for initial velocity. It's just not zero.
 
  • #19
Doc Al said:
D'oh! :wink:


The dropped object will have the initial speed of whatever it's being dropped from. (In this case from a moving elevator.)

Hi, but the speed of the elevator is going from the other direction? how does the upward speed transfer to a downward speed?
thanks
 
  • #20
emyt said:
Hi, but the speed of the elevator is going from the other direction? how does the upward speed transfer to a downward speed?
thanks
What do you mean? The initial speed of the bolt when it starts to fall from the elevator is the same as that of the elevator: 8 m/s upward. (After that, gravity exerts its influence, as for any projectile.)
 
  • #21
Doc Al said:
What do you mean? The initial speed of the bolt when it starts to fall from the elevator is the same as that of the elevator: 8 m/s upward. (After that, gravity exerts its influence, as for any projectile.)


I'm not sure I fully understand... the bolt is DROPPED, so its orientation is opposite to the upward speed of the elevator? I would be certain if the elevator and the bolt were going in the same direction...

thanks
 
  • #22
emyt said:
I'm not sure I fully understand... the bolt is DROPPED, so its orientation is opposite to the upward speed of the elevator?
Why opposite? It's dropped--not thrown or pushed.

Think of the example with the car. You're in a car, riding along at constant velocity. You drop something. Does that something start moving in the opposite direction? No, its initial velocity is the same as that of the car.
 
  • #23
Doc Al said:
Why opposite? It's dropped--not thrown or pushed.

Think of the example with the car. You're in a car, riding along at constant velocity. You drop something. Does that something start moving in the opposite direction? No, its initial velocity is the same as that of the car.


but being dropped is going down, so it's still opposite isn't it? since the elevator is going up

I can see your analogy but I'm having trouble accepting it (of course, that is entirely my fault). the claim seems to be "intuitive" but it isn't very intuitive to me..

when a speed is going in one direction, isn't there some kind of force that makes something go in that direction? if you drop something that isn't "going the same way", would it really have the same speed? perhaps I'm thinking of it like something going against the stream or a current?

sorry, but thanks for your patience
 
  • #24
emyt said:
but being dropped is going down, so it's still opposite isn't it? since the elevator is going up

I can see your analogy but I'm having trouble accepting it (of course, that is entirely my fault). the claim seems to be "intuitive" but it isn't very intuitive to me..

when a speed is going in one direction, isn't there some kind of force that makes something go in that direction? if you drop something that isn't "going the same way", would it really have the same speed? perhaps I'm thinking of it like something going against the stream or a current?

sorry, but thanks for your patience

Yes, for as long as the bolt was part of the elevator, it was being accelerated by the same force that was accelerating the rest of the elevator.

When something is dropped, it just means that it is no longer traveling as part of whatever platform it was dropped off of.

The bolt, still connected to the elevator, was accelerated for some time to the velocity of the elevator. 8 \tfrac{m}{s}

At that point, it broke off. What force, exactly, changed its velocity from 8 \tfrac{m}{s} to 0 \tfrac{m}{s}?
That's right, none.

And according to Newton's First Law:
An object in motion tends to stay in motion, and an object at rest tends to stay at rest unless acted upon by an external force.

With no force to "remove" its initial velocity (Since it was traveling along with the elevator), the velocity of the bolt remains the same as it was the instant it detached from the elevator.
Just like how an object you drop out of the window of a moving car, has an initial velocity that's the same as that of the car the moment it was dropped.
 
  • #25
emyt said:
but being dropped is going down, so it's still opposite isn't it? since the elevator is going up
No. Being dropped just means "let go". Usually you drop things from rest, so they start going down immediately. But not necessarily.

Imagine you are shot straight up into the air out of a circus cannon holding a baseball. As you are going up, you let the ball go. (Ignore air resistance.) What happens to the ball?

when a speed is going in one direction, isn't there some kind of force that makes something go in that direction?
No, why do you think that? (Note that force is only needed to change motion, not to maintain motion.)
if you drop something that isn't "going the same way", would it really have the same speed?
How can it not be "going the same way"? You're holding it! Once you let go, of course, other forces take over. But initially it's moving with the same speed and direction as you are. How could it not?
perhaps I'm thinking of it like something going against the stream or a current?
Perhaps. But even then, if you are holding something and let it go it starts out moving at the same speed and direction that you are moving.
 
  • #26
RoyalCat said:
Yes, for as long as the bolt was part of the elevator, it was being accelerated by the same force that was accelerating the rest of the elevator.

When something is dropped, it just means that it is no longer traveling as part of whatever platform it was dropped off of.

The bolt, still connected to the elevator, was accelerated for some time to the velocity of the elevator. 8 \tfrac{m}{s}

At that point, it broke off. What force, exactly, changed its velocity from 8 \tfrac{m}{s} to 0 \tfrac{m}{s}?
That's right, none.

And according to Newton's First Law:
An object in motion tends to stay in motion, and an object at rest tends to stay at rest unless acted upon by an external force.

With no force to "remove" its initial velocity (Since it was traveling along with the elevator), the velocity of the bolt remains the same as it was the instant it detached from the elevator.
Just like how an object you drop out of the window of a moving car, has an initial velocity that's the same as that of the car the moment it was dropped.

Thanks, I'm starting to accept it.
"At that point, it broke off. What force, exactly, changed its velocity from 8 to 0?" Is a very good point.

But aren't things like velocity associated with a direction? so following that thought, if something is let go, it would go in the same direction as the rest of the object.. but then I think about it, since gravity is now asserting itself, it would start going downwards instead of upwards.. so is there a point in time "instantaneously" perhaps? where the bolt is going in the same orientation as the elevator? (or is this what this whole initial velocity thing was all about)

thanks
 
  • #27
emyt said:
Thanks, I'm starting to accept it.
"At that point, it broke off. What force, exactly, changed its velocity from 8 to 0?" Is a very good point.

But aren't things like velocity associated with a direction? so following that thought, if something is let go, it would go in the same direction as the rest of the object.. but then I think about it, since gravity is now asserting itself, it would start going downwards instead of upwards.. so is there a point in time "instantaneously" perhaps? where the bolt is going in the same orientation as the elevator? (or is this what this whole initial velocity thing was all about)

thanks

Gravity is an acceleration, it changes the velocity. As time goes by, gravity makes the velocity of an object be more and more "pointed down," for it to instantly change the direction of a velocity, the acceleration due to gravity would have to be infinite in size.
 
  • #28
RoyalCat said:
Gravity is an acceleration, it changes the velocity. As time goes by, gravity makes the velocity of an object be more and more "pointed down," for it to instantly change the direction of a velocity, the acceleration due to gravity would have to be infinite in size.

okay, I understand it, it makes a lot of sense. thanks everybody

I just experimented for myself just now but raising an object into the air and letting go - it doesn't start going down right away hahah, that's very interesting
 
  • #29
emyt said:
okay, I understand it, it makes a lot of sense. thanks everybody

I just experimented for myself just now but raising an object into the air and letting go - it doesn't start going down right away hahah, that's very interesting

You've only just noticed? :-p

If it did, you wouldn't really be able to throw anything. :p
 
  • #30
RoyalCat said:
You've only just noticed? :-p

If it did, you wouldn't really be able to throw anything. :p

yes, it shows how much I've been paying attention :P

but on a more serious note, I've always understood the concept of throwing something - of course it would have the same initial velocity if it's going in the same direction. however, dropping something while it was previously going up felt different to me

thanks for all the help, I've been enlightened :P
 
  • #31
Hi, I'm having some trouble finishing the second part... here's the question again for quick reference

An elevator ascends with an upward acceleration of 4.0ft/s^2. At the instant its upward speed is 8.0ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor.

Calculate (a) the time of flight of the bolt from the ceiling to the floor

(b) the distance it has fallen relative to the elevator shaft.

Answer: (a) 0.71 (b) 2.3 ft

I supposed that at some point in time, the velocity is at 10.84, since vx = vx0 + axt --> vx = 8.0 + 4.0(0.71)--> 8.0 + 2.84 = 10.84

now since we're talking about the elevator shaft, I set x0 to zero and attempted to solve for x with various equations.. I can't seem to do it.. Could someone please give me some advice?

thanks

also, in one dimensional motion, if x = 0 vx0 would equal to 0 as well right? so we can say that Vx0 = 0?
 
  • #32
emyt said:
I supposed that at some point in time, the velocity is at 10.84, since vx = vx0 + axt --> vx = 8.0 + 4.0(0.71)--> 8.0 + 2.84 = 10.84
That's the velocity of the elevator at the moment the bolt hits the floor. You won't need that.

now since we're talking about the elevator shaft, I set x0 to zero and attempted to solve for x with various equations.. I can't seem to do it.. Could someone please give me some advice?
You use the same equations you used to solve part a. I would set the initial position of the bolt to be x0 = 9 and the initial position of the elevator floor to be x0 = 0.


also, in one dimensional motion, if x = 0 vx0 would equal to 0 as well right? so we can say that Vx0 = 0?
No, that's not true. You should know what Vx0 is by now! :wink:
 
  • #33
I don't understand. Why do you keep thinking that the initial velocity is zero?
 
  • #34
DaveC426913 said:
I don't understand. Why do you keep thinking that the initial velocity is zero?
I don't, after failing to find out how to solve the 2nd part I was wondering if x = 0 in one dimensional motion implied that the velocity is 0... since x/ time?

thanks
 
  • #35
Doc Al said:
That's the velocity of the elevator at the moment the bolt hits the floor. You won't need that.


You use the same equations you used to solve part a. I would set the initial position of the bolt to be x0 = 9 and the initial position of the elevator floor to be x0 = 0.



No, that's not true. You should know what Vx0 is by now! :wink:


thanks, damn. I thought I was on the right track - vx0 being 0 or not, I thought that if I could find the position of the bolt in the elevator shaft by setting x0 = 0 and finding the time it would take to get to the velocity of the elevator at that point..et c
 
  • #36
this is the full question:

An elevator ascends with an upward acceleration of 4.0ft/s^2. At the instant its upward speed is 8.0ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor.

Calculate (a) the time of flight of the bolt from the ceiling to the floor

(b) the distance it has fallen relative to the elevator shaft.

Answer: (a) 0.71 (b) 2.3 ft

Here I will solve (good practice for AP physics anyways)

So first let's do the time from dropping to the time where it returns to the original drop height...

first conversions:
8.0ft = 2.4384 meters
9.0ft= 2.7432 meters

so just do this

-2.7432 = -2.4384t + 1/2(-9.80)t^2

so

0 = (-4.9)t^2 + (-2.4384)t + 2.7432

so use quadratic formula

t = (2.4384 + sqrt(-2.4384^2 * 4(-4.9)(2.7432))) / (2*-4.9)

so i had t=.540 s
 
Last edited:
  • #37
bleedblue1234 said:
this is the full question:

An elevator ascends with an upward acceleration of 4.0ft/s^2. At the instant its upward speed is 8.0ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor.

Calculate (a) the time of flight of the bolt from the ceiling to the floor

(b) the distance it has fallen relative to the elevator shaft.

Answer: (a) 0.71 (b) 2.3 ft

Here I will solve (good practice for AP physics anyways)

So first let's do the time from dropping to the time where it returns to the original drop height...

first conversions:
8.0ft = 2.4384 meters
9.0ft= 2.7432 meters

so Vy = Vyo + gt

so -2.4384 = 2.4384 + (-9.80)t

so -4.8768 = (-9.80)t

so t = .4976326531 (i will store this in my calculator)

So now we know the time from dropping to the time where it is once against at the drop height...

so now we need to find the time from the original height to the floor...

(delta) y = Vyot + 1/2gt^2

2.7432 m = -4.8768t + 1/2(-9.8)t^2

so 0 = -4.9t^2 + -4.8768t + -2.7432 m

so use quadratic equation

t = (4.8768 - sqrt(-4.8768^2 - 4(-4.9)(-2.7432)))/2(-4.9)

so t=9.627132911 (store in calculator)

so if you add both t's you get 10.1 seconds
the question doesn't ask you for the time for the elevator shaft?
 
  • #38
emyt said:
the question doesn't ask you for the time for the elevator shaft?

?

i must have done my math wrong but i am sure the way to solve it is correct..
 
  • #39
bleedblue1234 said:
?

i must have done my math wrong but i am sure the way to solve it is correct..

what is 10.1 seconds supposed to be?
 
  • #40
i calculated the time from the bolt dropping to the time it hit the ground, but i think the correct answer is around .5 ish seconds...
 
  • #41
bleedblue1234 said:
i calculated the time from the bolt dropping to the time it hit the ground, but i think the correct answer is around .5 ish seconds...

you mean dropping to the floor of the elevator? that should be 0.71 seconds
 
  • #42
emyt said:
you mean dropping to the floor of the elevator? that should be 0.71 seconds

oooo i thought you were dropping from the top of the elevator to the bottom of the shaft... that's different
 
  • #43
bleedblue1234 said:
oooo i thought you were dropping from the top of the elevator to the bottom of the shaft... that's different

no, that's not the question.. how would you do what is actually stated there?

thanks
 
  • #44
Have you tried to draw a graph of the problem? I think a graph showing v/t might help you visualize the components better.
 
  • #45
DaveC426913 said:
Have you tried to draw a graph of the problem? I think a graph showing v/t might help you visualize the components better.

I haven't tried drawing a graph but I've already drawn graphs to visualize the components of a velocity vector. but this problem has motion in 1 dimension right? so it's just a graph of x versus time?

thanks
 
  • #46
emyt said:
I haven't tried drawing a graph but I've already drawn graphs to visualize the components of a velocity vector. but this problem has motion in 1 dimension right? so it's just a graph of x versus time?

thanks

Yes.
 
  • #47
What am I doing wrong in this graph?

I'm having trouble figuring how to skew the bolt drop slope to account for the initial velocity.

I've skewed it by y=1 across x=1. (faint pink line almost visible to left of red line)


Clearly I've got it very wrong since it shows a fall lasting only .25s.
 

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  • #48
emyt said:
thanks, damn. I thought I was on the right track - vx0 being 0 or not, I thought that if I could find the position of the bolt in the elevator shaft by setting x0 = 0 and finding the time it would take to get to the velocity of the elevator at that point..et c
Start by showing how you solved part a. What you should have done to solve it, is equate expressions for the position of the bolt and the floor of the elevator (with respect to the shaft or ground) as a function of time. You solve for the time when both are at the same place. Then just plug that time into get the position.
 
  • #49
bleedblue1234 said:
Here I will solve (good practice for AP physics anyways)

So first let's do the time from dropping to the time where it returns to the original drop height...

first conversions:
8.0ft = 2.4384 meters
9.0ft= 2.7432 meters

so just do this

-2.7432 = -2.4384t + 1/2(-9.80)t^2

so

0 = (-4.9)t^2 + (-2.4384)t + 2.7432

so use quadratic formula

t = (2.4384 + sqrt(-2.4384^2 * 4(-4.9)(2.7432))) / (2*-4.9)

so i had t=.540 s
Please do not (attempt to) provide complete solutions. Let the OP figure it out!
 
  • #50
DaveC426913 said:
What am I doing wrong in this graph?
You show a speed vs time graph, yet the top and bottom of the elevator car are on different lines. (I think we can safely assume that they have the same speed!)

Draw a position vs time graph.
 

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