Register to reply 
A Numerical on motion in one dimension . 
Share this thread: 
#1
Jun1911, 04:11 AM

P: 758

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(vu) = 1/2v If time = t seconds then of course t=t2 So s=s2=1/2v 


#2
Jun1911, 05:39 AM

Mentor
P: 41,440

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
Jun1911, 09:37 AM

P: 758




#4
Jun1911, 09:45 AM

Mentor
P: 41,440

A Numerical on motion in one dimension .



#5
Jun1911, 11:43 AM

P: 758




#7
Jun2011, 10:47 AM

P: 758

please download it . Since ball is vertically thrown, it should not be called 'projectile'. 


#8
Jun2011, 11:28 AM

Mentor
P: 41,440

(FYI: I cannot access that site from my office, but I'll look at it later.) 


#9
Jun2011, 01:25 PM

P: 758

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
Jun2011, 01:50 PM

Mentor
P: 41,440

y = y_{0} + v_{0}t  1/2gt^{2} Where v_{0} 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
Jun2011, 01:53 PM

P: 758

This means displacement of both the balls will be different according to you , right ?? 


#12
Jun2011, 02:04 PM

Mentor
P: 41,440




#13
Jun2011, 02:09 PM

P: 758

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 ??) 


#14
Jun2011, 02:20 PM

Mentor
P: 41,440

y = vt  1/2gt^{2} y = vt  1/2gt^{2} 


Register to reply 
Related Discussions  
Projectile motion numerical help needed  Advanced Physics Homework  4  
Motion in One Dimension  Introductory Physics Homework  3  
Projectile motion related numerical  Introductory Physics Homework  4  
Motion in One Dimension  Introductory Physics Homework  1  
Motion in one dimension  Introductory Physics Homework  2 