A ball is dropped from rest from the top of a building of
height h. At the same instant, a second ball is projected vertically
upward from ground level, such that it has zero speed when it
reaches the top of the building.
(A) At the time when the two balls pass each other –
which ball has the greater speed, or do they have the same speed?
(B) Where will the two balls be when they are alongside each
other: at height h/2 above the ground, below this height, or above
2. Relevant equations
Equations of 1D motion for a constant acceleration of –g
3. The attempt at a solution
First here is the diagram I drew:
Now for part (A):
I suppose that the second ball was thrown up with speed –v0
And the balls pass each other at time t
So for the first ball v = -gt
And for the second ball v = -v0 -gt
But we are concerned with speed, so I shall ignore the signs, right?
Then certainly gt < v + gt
Thus when the balls meet the second ball has higher speed…. Am I correct?
For part (B):
I think the balls will be alongside each other when they have same position (y value).
So for the first ball, y = -0.5 gt2
And for the second ball, y-(-h) = -v0t -0.5gt2
Now, here I have two confusions:
1. Shouldn’t I write –y = -0.5 gt2 ? Because the ball is certainly below y = 0.
2. Is that above equation for the second ball is ok? I used y-(-h) instead of y because I thought the initial position of ball 2 is –h and thus the displacement is y-(-h). Is this ok?
I shall go for rest of my solution after someone ensures my approach is correct (or is not).
Thanks in advance!