A Numerical on motion in one dimension .

In summary: I don't understand this. The initial speed is -v, not zero.As per the formula I gave before:y...As I stated before, yes.In summary, The position as a function of time for a ball thrown upwards and a ball thrown downwards with the same initial velocity v will be different due to the difference in initial velocity. The ball thrown upwards will reach its highest point with a velocity of 0, while the ball thrown downwards will start with a velocity of -v. Therefore, their displacements will also be different.
  • #1
sankalpmittal
785
15

Homework Statement



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 ?

Homework Equations



According to me the equations of motions are relevant here .

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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  • #2
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?
 
  • #3
Doc Al said:
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 .
 
  • #4
sankalpmittal said:
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.
 
  • #5
Doc Al said:
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 .
 
  • #6
sankalpmittal said:
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.
 
  • #8
sankalpmittal said:
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.)
 
  • #9
Doc Al said:
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 .
 
  • #10
sankalpmittal said:
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.
 
  • #11
Doc Al said:
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 ??
 
  • #12
sankalpmittal said:
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.
 
  • #13
Doc Al said:
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
= 1/2gt2^2Therefore s1 is not equal to s2 right ??
(Displacements of two balls are different right ??)
 
  • #14
sankalpmittal said:
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.
 
  • #15
Doc Al said:
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 .

:cool:
 

1. What is motion in one dimension?

Motion in one dimension refers to the movement of an object along a straight line. It is also known as linear motion and can be described using distance, displacement, speed, velocity, and acceleration.

2. How is velocity calculated in one-dimensional motion?

Velocity is the rate of change of an object's displacement. It is calculated by dividing the change in displacement by the change in time. In one-dimensional motion, velocity is a vector quantity with both magnitude and direction.

3. What is the difference between distance and displacement in one-dimensional motion?

Distance is the total length of the path an object has traveled, while displacement is the straight-line distance between an object's initial and final position. Distance is a scalar quantity, while displacement is a vector quantity.

4. How is acceleration related to velocity in one-dimensional motion?

Acceleration is the rate of change of an object's velocity. It can be calculated by dividing the change in velocity by the change in time. In one-dimensional motion, acceleration can either be positive (speeding up) or negative (slowing down).

5. What is the difference between uniform and non-uniform motion in one dimension?

Uniform motion is when an object moves with a constant velocity, meaning its speed and direction do not change. Non-uniform motion is when an object's velocity is changing, either by changing speed or direction, or both.

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