# 2 balls dropped from a building

1. Nov 5, 2007

### ronny45

1. The problem statement, all variables and given/known data
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.00s later. You may ignore air resistance.
If the height of the building is 21.0m, what must the initial speed be of the first ball if both are to hit the ground at the same time?

2. Relevant equations
v=u+at
v^2=u^2+2as
2s=ut+1/2 at^2

3. The attempt at a solution

For the second ball,
u=0
a=9.81m/s^2
s=21m
v=?
....v comes out to be 20.298m. Then i found t to be 2.069s. Assuming the first ball was in the air 1 second longer, its time is 3.069.

However, I' not sure how to proceed with the equations for the first ball, since I don't know the maximum height reached by it, though v is 0 at this point.

2. Nov 5, 2007

### gknowels

since you know the time, acceleration, initial and final positions, you should be able to use x = x_0 + v_0 t + (1/2) a t^2 to find the answer.