How Do Two Balls Thrown at Different Speeds and Heights Meet?

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To determine when two balls thrown at different speeds and heights meet, the equations of motion must be set equal, as both distance and time are shared variables. The first ball is thrown upward from the ground at 25 m/s, while the second ball is dropped from a 15 m high building. The relevant equation for both is D = Vi(t) + 1/2 at^2. A side discussion arose regarding the implications of using imaginary numbers in physics equations, particularly how they affect outcomes related to time. Understanding these concepts is crucial for solving the problem accurately and preparing for related test questions.
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1. A ball is thrown upward from the ground with an initial speed of 25m/s. At the same instant, a ball is dropped from rest from a building 15m high. After how long with the balls be at the same height?
2. The equation i had in mind was D=Vi(t) + 1/2 at^2
3. For this problem I believe that I have to set two sets of the above equation equal to each other since distance or time for both of the balls have to be equal. However I do not know how to do this exactly.

edit: I just realized there was another post on this problem, however i could not find help on it. Also as a side question, my physics teacher said a bonus question on our upcoming test will be what happens to all the outcomes of equations including time if time is an imaginary number. I told him that they would become zero or decrease or become negative and said i was wrong. Any thoughts about this? To me, an imaginary number is the square root of a negative number.
 
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Almost - that's probably the correct equation
Remember that the distance for one of the balls is (15-D) depending on which level you want to use as zero and note that a is the same sign for both.
 
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