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Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. What is the difference in the time the balls spend in the air? What is the velocity of each ball as it strikes the ground? How far apart are the balls .800 s after they are thrown.

I can't figure out how to set up this equation. I would think it has to use the displacement with constant acceleration equation (delta y = initial velocity * delta t + .5accl * delta t^2), but it has two variables unsolved for (displacement & time)... help, please?

Please help????

I can't get this on my own. I've been trying.

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