Motion of two balls which collide

  • Thread starter Faka
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  • #1
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A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. How to I find the time at which the two balls collide. A how do I find the value of H in terms of v0 and g so that at the instant when the balls colide, the first ball is at the heighest point of its motion.

Anybody who can help?
 

Answers and Replies

  • #3
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Is it this equation I need to use s = u*t + 1/2*a*t^2 ?
 
  • #4
tiny-tim
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Yes. :smile:
 
  • #5
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Ball 1 (Going upward with a v0):

s1 = vo*t-0,5*g*t^2

Ball 2 (Dropped from a height H)

s2 = h-0,5*g*t

And then I need to find the time t:
s1 = s2

I get something crazy !

Is my method good?
 
  • #6
tiny-tim
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Hi Faka! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
Ball 2 (Dropped from a height H)

s2 = h-0,5*g*t

you mean s2 = h - 0,5*g*t2 :wink:

otherwise, that's ok :smile:
 
  • #7
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Yes, its t^2 .
I think that I have made a mistake. Shouldn't it be:
Ball 1

s1 = Vo*t+0,5*g*t^2

Ball 2

s2 = h+0,5*g*t

Then
s1 = s2

I mean "+" instead of "-".
 
  • #8
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s2 = h+0,5*g*t^2
 
  • #9
tiny-tim
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it doesn't matter whether you use +g and g = -9.81, or -g and g = -9.81 :wink:

show us your full calculations :smile:
 
  • #10
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Ball 1 (Upward)
x1 = 0 + v0*t + 1/2*(g)*t^2
= vo*t + g*t^2

Ball 2 (Dropped)
x2 = h + 0*t + 1/2*(-g)*t^2
= h - 1/2*g*t^2


x1 = x2

vo*t + g*t^2 = h - 1/2*g*t^2


And then I solve for t, I get something that I cant write here. Its something divided by 2*g
 
  • #11
tiny-tim
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Ball 1 (Upward)
x1 = 0 + v0*t + 1/2*(g)*t^2
= vo*t + g*t^2

Ball 2 (Dropped)
x2 = h + 0*t + 1/2*(-g)*t^2
= h - 1/2*g*t^2

No, you can't have g in one equation and -g in the other. :redface:
 

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