# Bob drops some balls.

1. Oct 4, 2012

### pmd28

Bob has two balls. He releases one ball from a platform of height H. Just as that ball strikes the ground he releases the next ball from the same height H. Assuming that the first ball bounces perfectly, (i.e. reversing only the direction of its velocity when it strikes the ground), at what height from the ground do the balls strike each other?
a) H/4
b) h/2
c) h/3
d) 3H/4
e) none of the above

the answer is d

I thought that when the first ball strikes the ground it has a momentary velocity of zero. And since it bounces "perfectly" it should return with the same velocity it was released with therefore the ball that was released would have the same kinematic properties. ergo they should meet in the middle. I'm so confused.

Last edited: Oct 4, 2012
2. Oct 4, 2012

### azizlwl

Use vertlcal motion for both downward and upward direction balls.
2 equations 2 unknown, y and t.

Last edited: Oct 4, 2012
3. Oct 4, 2012

### pmd28

So for ball 1 (the ball that is returning up) my variables are:
v= (dunno, I think they would be the same for both.)
v(initial)= 0
t=?
a=g
y=?

and variables for ball 2 are the same. And when I solve the equation i still get H/2

4. Oct 12, 2012

### azizlwl

It is a 1D motion
One from top with y0=H, v0=0
On from bottom, with kinetic energy of equal mgh. You can convert this to v0.

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