Im having trouble with this problemPlease help

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To determine the speed of the second ball thrown upward so that it meets the first ball dropped from height h at h/2, the time taken for the first ball to fall can be calculated using the equation h/2 = 1/2gt^2, leading to t = (h/g)^(1/2). Since both balls meet at the same height and time, the second ball's upward motion must be analyzed using its initial speed and the same time t. The discussion emphasizes that the time for both balls is equal, which is crucial for solving the problem. Understanding the relationship between time, distance, and acceleration is key to finding the correct speed for the second ball.
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A ball is dropped from rest a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.

Ive been working so long on homework today maybe that's the reason I can't figure this one out...can someone please help me Thankyou
 
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What have you done so far?
 
For the ball released from height h, h/2 = 1/2gt^2, where t is the time taken to fall a distance of h/2. Therefore t = (h/g)^1/2...i have no clue where to go next or if this is even correct
 
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You can almost intuitively even figure this one out. You seem to be off to a good start though. Also, remember that you know that the time is the same for both since they meet at the same time.
 
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