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

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The discussion centers on the physics problem of determining when two balls, one thrown upward at 25 m/s and the other dropped from a height of 15 m, will meet at the same height. The relevant equation used is D = Vi(t) + 1/2 at², where D represents distance, Vi is the initial velocity, t is time, and a is acceleration due to gravity. Participants emphasize the need to set the distance equations for both balls equal to each other to find the time at which they meet. Additionally, a side question regarding the implications of imaginary numbers in physics equations was raised, highlighting the complexity of time as a variable.

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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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