2 balls dropped from a building

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SUMMARY

The problem involves two balls: one thrown upwards from a 21.0m high building and another dropped 1.00s later. To determine the initial speed of the first ball so that both hit the ground simultaneously, the equations of motion are applied. The second ball, dropped with an initial velocity of 0 and under the acceleration of gravity (9.81 m/s²), takes 2.069 seconds to reach the ground. Consequently, the first ball must be in the air for 3.069 seconds, requiring the use of the kinematic equation x = x₀ + v₀t + (1/2)at² to find the necessary initial speed.

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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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