Ball Release and Thrown to reach ground at the same time

[spikehoward]

Homework Statement

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00s later. You may ignore air resistance. If the height of the building is 20m, what must the initial speed of the first ball be if both are to hit the ground at the same time?

Homework Equations

Vf=Vo+a*t
X=Vo*t+1/2*a*t^2

The Attempt at a Solution

Ball 1:
X=Vo*t+1/2*a*t^2
20=0+1/2*9.8*t^2
t=2.02s

Ball 2:
total time=1+2.02=3.02

The ball is thrown up, eventually comes to a rest, and then begins to fall so
Vf=0=Vo-g*t1 so Vo=g*t1

where t1 is the time it takes for the ball to reach its peak AND the time it will take to fall back to it original position. The velocity, Vo has the same magnitude and opposite direction and the point it is initially thrown and when it returns.

(1) Vo=g*t1
(2)20=Vo*t2+1/2*g*t2^2
(3) 2*t1+t2=3 where t2 is the time it takes to fall from the initial point to the bottom.

Can I use these three equation to solve? I get an answer that checks out but the solution manual says it is incorrect. Did I make a wrong assumption?

 

[TSny]

I think your equations should give the correct solution. I got a little confused at first because "Ball 1" in your solution is the "second ball" in the problem statement. But that's OK.

You are apparently taking downward as positive direction.

For the ball that is thrown upward, you don't have to break up the flight into parts. You can treat the entire flight in one equation. So, you don't need to deal with t1 and t2 at all. You just need the total time (3.02 s).