2 balls dropped from a building

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To determine the initial speed of the first ball thrown from a 21.0m building so that both balls hit the ground simultaneously, the second ball's parameters are established with an initial velocity of 0 m/s and an acceleration of 9.81 m/s², resulting in a fall time of 2.069 seconds. The first ball, thrown upwards, must remain in the air for 3.069 seconds, accounting for the additional second of flight. The equations of motion can be applied to find the initial velocity of the first ball, although the maximum height it reaches remains unknown. The key is to utilize the time of flight and the equations of motion to solve for the initial speed. The calculations will yield the necessary initial speed for the first ball to ensure both balls reach the ground simultaneously.
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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 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?

Homework Equations


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


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.
 
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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.
 
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