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A Numerical on motion in one dimension .

by sankalpmittal
Tags: dimension, motion, numerical
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sankalpmittal
#1
Jun19-11, 04:11 AM
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1. The problem statement, all variables and given/known data

A boy throws a ball with v velocity upwards as well as downwards . What will be the displacement covered by the two balls if air friction and external force or viscosity is neglected ?

2. Relevant equations

According to me the equations of motions are relevant here .

3. The attempt at a solution

ummm
I case :

u=0
t= 1 sec to gain velocity v
v= v

then s(displacement) = 1/2t(v+u)
= 1/2v

II case :
u2=0
t2= 1 sec to gain velocity v
v= v
then s2(displacement) = 1/2t(v-u)
= 1/2v
If time = t seconds then of course t=t2

So s=s2=1/2v
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Doc Al
#2
Jun19-11, 05:39 AM
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I don't quite understand what you're doing here. Does u stand for the initial velocity? How can it be zero if you're throwing things?
sankalpmittal
#3
Jun19-11, 09:37 AM
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Quote Quote by Doc Al View Post
I don't quite understand what you're doing here. Does u stand for the initial velocity? How can it be zero if you're throwing things?
You hold ball in your hand , so it is at rest . Then you throw , so it gains velocity v in 1 second . Yes u stand for initial velocity .

Doc Al
#4
Jun19-11, 09:45 AM
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A Numerical on motion in one dimension .

Quote Quote by sankalpmittal View Post
You hold ball in your hand , so it is at rest . Then you throw , so it gains velocity v in 1 second . Yes u stand for initial velocity .
Are you trying to analyze the motion of the ball while it's in your hand or after you've throw it? Once you've thrown it and it leaves your hand, it's just a projectile.
sankalpmittal
#5
Jun19-11, 11:43 AM
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Quote Quote by Doc Al View Post
Are you trying to analyze the motion of the ball while it's in your hand or after you've throw it? Once you've thrown it and it leaves your hand, it's just a projectile.
No the ball is thrown upwards in one direction and it makes the angle of 90 degree , its not a projectile .
Doc Al
#6
Jun19-11, 11:51 AM
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Quote Quote by sankalpmittal View Post
No the ball is thrown upwards in one direction and it makes the angle of 90 degree , its not a projectile .
Once it leaves your hand it is.
sankalpmittal
#7
Jun20-11, 10:47 AM
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Quote Quote by Doc Al View Post
Once it leaves your hand it is.
http://www.filefactory.com/file/cce8f02/n/untitled.bmp


please download it . Since ball is vertically thrown, it should not be called 'projectile'.
Doc Al
#8
Jun20-11, 11:28 AM
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Quote Quote by sankalpmittal View Post
Since ball is vertically thrown, it should not be called 'projectile'.
It doesn't matter how you throw it--up, down, sideways--as soon as it leaves your hand it is a projectile. (Even if it moves only vertically, it's still a projectile. A simple case, since you don't have any horizontal motion to worry about.)

(FYI: I cannot access that site from my office, but I'll look at it later.)
sankalpmittal
#9
Jun20-11, 01:25 PM
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Quote Quote by Doc Al View Post
It doesn't matter how you throw it--up, down, sideways--as soon as it leaves your hand it is a projectile. (Even if it moves only vertically, it's still a projectile. A simple case, since you don't have any horizontal motion to worry about.)

(FYI: I cannot access that site from my office, but I'll look at it later.)
Ok , I searched on the net , you are correct . But what about the numerical :

A boy throws a ball with v velocity upwards and another ball with same velocity downwards . What will be the displacement covered by the two balls if air friction and external force or viscosity is neglected ?

What will be the answer ???


Thanks in advance .

Please do download the image whenever possible .
Doc Al
#10
Jun20-11, 01:50 PM
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Quote Quote by sankalpmittal View Post
Ok , I searched on the net , you are correct . But what about the numerical :

A boy throws a ball with v velocity upwards and another ball with same velocity downwards . What will be the displacement covered by the two balls if air friction and external force or viscosity is neglected ?

What will be the answer ???
If you want the vertical position as a function of time, use the kinematic formula:
y = y0 + v0t - 1/2gt2

Where v0 is the initial velocity. In one case the initial velocity will be +v; in the other, -v. That's the only difference between the two cases.
sankalpmittal
#11
Jun20-11, 01:53 PM
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Quote Quote by Doc Al View Post
If you want the vertical position as a function of time, use the kinematic formula:
y = y0 + v0t - 1/2gt2

Where v0 is the initial velocity. In one case the initial velocity will be +v; in the other, -v. That's the only difference between the two cases.
Ok so you mean then in one case velocity is + while in other its - .

This means displacement of both the balls will be different according to you , right ??
Doc Al
#12
Jun20-11, 02:04 PM
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Quote Quote by sankalpmittal View Post
Ok so you mean then in one case velocity is + while in other its - .
Yes.
This means displacement of both the balls will be different according to you , right ??
The position as a function of time will be different for each case. The ball thrown upward first has to reach the highest point, then return to the starting position. At that point, it's exactly the same as the other ball started out--it's velocity is v downward.
sankalpmittal
#13
Jun20-11, 02:09 PM
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Quote Quote by Doc Al View Post
Yes.

The position as a function of time will be different for each case. The ball thrown upward first has to reach the highest point, then return to the starting position. At that point, it's exactly the same as the other ball started out--it's velocity is v downward.
Ist Case : Throwing upwards

u=v
v=0 (on reaching highest point)

s1 = ut - 1/2gt1^2


2nd Case : Throwing downwards

u=0
v=v

s2 = ut2+ 1/2gt2^2
= 1/2gt2^2


Therefore s1 is not equal to s2 right ??
(Displacements of two balls are different right ??)
Doc Al
#14
Jun20-11, 02:20 PM
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Quote Quote by sankalpmittal View Post
Ist Case : Throwing upwards

u=v
v=0 (on reaching highest point)
The initial velocity is +v. Why mention the velocity at the highest point?

s1 = ut - 1/2gt1^2
As per the formula I gave before:
y = vt - 1/2gt2


2nd Case : Throwing downwards

u=0
v=v
I don't understand this. The initial speed is -v, not zero.

s2 = ut2+ 1/2gt2^2
= 1/2gt2^2
As per the formula I gave before:
y = -vt - 1/2gt2


Therefore s1 is not equal to s2 right ??
(Displacements of two balls are different right ??)
As I stated before, yes.
sankalpmittal
#15
Jun20-11, 10:56 PM
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Quote Quote by Doc Al View Post
The initial velocity is +v. Why mention the velocity at the highest point?


As per the formula I gave before:
y = vt - 1/2gt2



I don't understand this. The initial speed is -v, not zero.


As per the formula I gave before:
y = -vt - 1/2gt2



As I stated before, yes.

OK .



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