Ball thrown upward and another is dropped

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To determine when the two balls will be at the same height, the motion of each ball must be analyzed using kinematic equations. The ball thrown upward will follow a trajectory defined by its initial speed of 25 m/s, while the dropped ball will fall under the influence of gravity from a height of 15 m. The equations of motion can be set up for both balls, allowing for the calculation of the time it takes for them to reach the same height. Visual aids, like diagrams, can help clarify the forces acting on each ball. Ultimately, solving the equations will yield the time at which both balls are at the same height.
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A ball is thrown upward from the ground with a initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height?

I'm not even sure how to start it so help!
 
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Draw a picture, with the applied forces.
 
First, consider just one ball.
Can you determine its motion as a function of time?
 
I'm not sure what formula to use to determine the time and I drew a pic but it didn't really help.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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