Throwing a ball straight up

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  • #1
chawki
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Homework Statement


A ball is thrown straight up and it falls back to the ground.


Homework Equations


Which one of the
following graphical representations describes the velocity of the ball as a
function of time.


The Attempt at a Solution


I think the right answer is D, However i think it can also be C.
Tell me what you think!
 

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Answers and Replies

  • #2
Doc Al
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How does the acceleration due to gravity relate to these diagrams?
 
  • #3
tiny-tim
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I think the right answer is D, However i think it can also be C.
Tell me what you think!

well, it can't be both, can it? :biggrin:

come on, chawki, you know the rules by now :wink:

you tell us why you think it's D (or why it might be C), and then we'll comment :smile:
 
  • #4
chawki
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Well, D because when we throw the ball, it will reach a time when the velocity will come to zero, and then it will fall, so basically there are two steps..
Step1: a1 < 0 and v1 > 0, a1*v1 < 0, so there is decceleration.
Step2: a2 < 0 and v2 < 0, a2*v2 > 0, so there is acceleration.

same thing for graph C
 
  • #5
tiny-tim
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Well, D because when we throw the ball, it will reach a time when the velocity will come to zero, and then it will fall, so basically there are two steps..
Step1: a1 < 0 and v1 > 0, a1*v1 < 0, so there is decceleration.
Step2: a2 < 0 and v2 < 0, a2*v2 > 0, so there is acceleration.

same thing for graph C

(try using the X2 icon just above the Reply box :smile:)

i'm confused :redface:

what is a1*v1 supposed to be? :confused:
 
  • #6
chawki
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a1 is the acceleration during first step
v1 is the velocity during first step
 
  • #7
tiny-tim
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but why are you multiplying them? :confused:
 
  • #8
Doc Al
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These diagrams show velocity as a function of time. Using them, how would you find the acceleration? Which diagram matches what you (should) know about the acceleration of an object in free fall?
 
  • #9
chawki
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but why are you multiplying them? :confused:

It's a method i learnt at college
 
  • #10
tiny-tim
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  • #11
chawki
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These diagrams show velocity as a function of time. Using them, how would you find the acceleration? Which diagram matches what you (should) know about the acceleration of an object in free fall?

The acceleration is simply a1= difference of velocity during that step/difference of time during that step
 
  • #12
chawki
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explain :confused:

a*v < 0 means there is decceleration
a*v > 0 means there is acceleration
 
  • #13
tiny-tim
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a*v < 0 means there is decceleration
a*v > 0 means there is acceleration

hmm … i find that rather confusing …

by "acceleration" we usually mean acceleration in a particular direction (eg up), so that the acceleration of a projectile is always negative

so the test is:
a < 0 means there is negative acceleration
a > 0 means there is positive acceleration​

your method seems to be defining "acceleration" as meaning that the magnitude of the velocity increases (so a projectile always decelerates on the way up and always accelerates on the way down)

i don't like your method at all, it's so complicated

surely a < 0 is an easier test than a*v < 0 ? :smile:

oh, and i still don't understand why, even on your test, you think C might be right :confused:
 
  • #14
chawki
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Yes C might be right too because when we apply that method which you don't like :D we find same thing as D
one other thing..i learnt in french..maybe they gave us different definitions..all i know and remember is that it works.
so what do you think C or D :D
 
  • #15
Doc Al
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The acceleration is simply a1= difference of velocity during that step/difference of time during that step
Exactly right. So how do the accelerations exhibited by diagrams C and D differ? Hint: Pay attention to signs.
 
  • #16
chawki
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if we take the method of looking only to the acceleration, i think the answer would be C only.
 
  • #17
Doc Al
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if we take the method of looking only to the acceleration, i think the answer would be C only.
Why do you say that?

What property of a velocity versus time graph tells you the acceleration?

Realize that acceleration is a vector and thus has direction (and thus sign).
 
  • #18
chawki
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yes yes, i'm aware.
i just used to use acceleration*velocity to see if it's acceleration or deceleration.
maybe i was wrong???
 
  • #19
tiny-tim
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hi chawki! :smile:

now i've slept on it :zzz:, i can see why your method is giving you C as an alternative answer

when the projectile comes down, it is accelerating (ie the magnitude of the velocity is increasing), so your method requires a*v to be positive

in the graph C, a and v are both positive, so from that point of view C is consistent with that

(and so is D, the correct answer, with both a and v negative)

unfortunately, C corresponds to half a parabola on the way up, then the other half of the parabola upside down, with the projectile accelerating off into outer space :rolleyes: (so that the whole thing looks a bit like a y = x3 graph)

(can you se that now?)

i hope you now see that this acceleration/deceleration method, using the sign of both a and v, is potentially misleading, and it's best just to talk about positive and negative acceleration, and use only the sign of a :smile:
 
  • #20
chawki
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Ok Tiny-tim,
it's a good point to clear up this confusing matter, i will use only 'a' to see if there is acceleration or deceleration, with a = difference in velocity/ difference in time.
and then the answer would be only C.
 
  • #21
tiny-tim
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… i will use only 'a' to see if there is acceleration or deceleration, with a = difference in velocity/ difference in time.
and then the answer would be only C.

why C ? :confused:

in which direction are you measuring v and a ?
 
  • #22
chawki
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because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.
 
  • #23
Doc Al
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because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.
What's step 1 and step 2? Are you still looking at diagram C?

Is the acceleration of a falling body sometimes positive and sometimes negative? (Which way does gravity act?)
 
  • #24
tiny-tim
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because
in step 1
a1= (0-v)/(t1-t0) and we get a negative value.
in step2
a2= (v-0)/(t2-t1) = positive value.

so it makes sense.

chawki, what you are saying (correctly) is that in step 1 of graph C, the acceleration upward is negative, and in step 2 of graph C, the acceleration upward is positive …

as Doc Al :smile: is hinting, do you really think that is correct for a projectile under gravity? :redface:
 
  • #25
Vulgar
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I don't mean to confuse chawki more, but is that graph a good representation? If they mean the size of the vector with v, can't it be C then? The size only isn't sign dependant right? Or do they mean the z-component of the velocity vector?
 
  • #26
chawki
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chawki, what you are saying (correctly) is that in step 1 of graph C, the acceleration upward is negative, and in step 2 of graph C, the acceleration upward is positive …

as Doc Al :smile: is hinting, do you really think that is correct for a projectile under gravity? :redface:

In step2 the acceleration is DOWNWARD and it's positive.
it makes sense because when we throw upward , a is Negative, and then the ball reach a pooint where its velocity comes to zero...before falling free and then the accleration is Positive...and all this matches C...still convinced :shy:
 

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  • #27
Doc Al
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In step2 the acceleration is DOWNWARD and it's positive.
What sign are you using to represent downward? Those diagrams assume that up is positive and down is negative.
it makes sense because when we throw upward , a is Negative, and then the ball reach a pooint where its velocity comes to zero...before falling free and then the accleration is Positive...and all this matches C...still convinced
Don't just think in terms of the magnitude of velocity, direction counts.

Answer this: What does a positive (> 0) value for v mean in those diagrams?
 
  • #28
chawki
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What sign are you using to represent downward? Those diagrams assume that up is positive and down is negative.

Don't just think in terms of the magnitude of velocity, direction counts.

Answer this: What does a positive (> 0) value for v mean in those diagrams?

I'm using downward as (+) upward as (-)
 
  • #29
chawki
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no matter what the methods are..we should compare results to logic.
both C and D are correct.
 
  • #30
Wolf5370
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Chawki I think you are confusing yourself with acceleration - ignore it for a moment and think only as velocity...

The speed of the ball (not velocity) starts off greatest just after the throw, reaches the apex where speed is zero, and drops back under gravity. So, sped (again not velocity) would show C.

Now velocity is not scalar, it has a direction component. As the ball is launched, gravity is slowing it, so it drops in a consistant manor (in a perfect example) over time until it meaches zero. Step 1 (?). Then it falls, but now the direction is reversed, so as the magnitude rises again (its speed increases), it gets more negative - thus D is the answer.
 
  • #31
Wolf5370
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In step2 the acceleration is DOWNWARD and it's positive.
it makes sense because when we throw upward , a is Negative, and then the ball reach a pooint where its velocity comes to zero...before falling free and then the accleration is Positive...and all this matches C...still convinced :shy:

No - gravity is the accelerator here - and is always in the same direction. It does not get faster when leaving the thrower's hand it decelerates and then when gravity wins, it accelerates - but ALWAYS in the downward direction. The ball has no motor or jet pack, it leaves with the force of the throw and starts to immediately decelerate under gravity, -ve (if you like) acceleartion until its KE is exhausted (now G.PE) and it drops once more.
 
  • #32
chawki
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If my method was wrong, i wouldn't find that D can be true too...
 
  • #33
Wolf5370
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If your method was correct, you would not find that C can be correct too! - Sorry.
 
  • #34
chawki
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If your method was correct, you would not find that C can be correct too! - Sorry.

C can be correct...the ball is thrown upward with an initial speed...at some time it reaches 0, ball change direction (velocity also) and head down to the ground, earning speed again.
 
  • #35
Doc Al
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I'm using downward as (+) upward as (-)
Well, you'd better think again. Use upward as (+) and downward as (-). Otherwise you're saying that when you throw a ball up, it's velocity is down. (Note the sign of the velocity on the diagrams!)

OK. Now look at diagram C again. What does it tell you? Does the object ever come down?
 

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