Do Two Balls Thrown at Different Times from a Cliff Pass Each Other?

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The discussion revolves around solving an antiderivative problem involving two balls thrown from a cliff, with one thrown a second after the other. Participants highlight the need to account for the time difference between the two balls, suggesting the use of separate time variables for accurate calculations. The key point is to set the positions of both balls equal to determine if they pass each other. An error was identified in the initial evaluation of the time difference, which was corrected in the later stages of the discussion. The solution emphasizes the importance of correctly applying the antiderivative to find the heights over time.
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Need Help With Antiderivative Problem!

We've put 3 people's minds to this, and every time we end up with 2 non-equal constants having to be equal for the equation to be true. Give it a whirl, b/c I have nothing left to lose... my dignity has already been shot by now.

Okay, here's the deal: 2 balls are thrown upward from the edge of a cliff 432 feet above the ground. The first is thrown with a speed of 48 f/s and the other is thrown a second later at 24 f/s. Do the balls ever pass each other? Note, other pertinent information is: we are following the work in the book which states: The motion is vertical and we choose the positive direction to be upward. At time T the distance above the ground is s(T) and the velocity v(T) is decreasing. Therefore the accelleration must be negative and we have a(T) = dv/dT = -32 (32 being the gravitational force). Thus the antiderivative v(T) = -32T + C This much should work for both the first and second balls. Then, given that v(0) = 48 {for the first ball} we see that v(T) = -32T +48 [Which becomes v(0) = 24 for the second ball, and v(T) = -32T + 24 for the second ball].

Now, we know the heights have to be equal in order for them to "pass" each other, but the problems we encounter are: The second ball is thrown ONE SECOND later than the first, and the second antidifferentiation of v(T) gives us the formula for the maximum height reached, not any height over a period of time T. SOMEONE PLEASE HELP!

Thanks in advance!
 
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You will need to keep a separate time variable for each ball to get your equations. Then let t2= t1-1

This will give you the position of the second ball in the same time variable as the first. Set the position of each ball expressed in terms of t1 equal to find if they pass.
 
Integral said:
You will need to keep a separate time variable for each ball to get your equations. Then let t2= t1-1

This will give you the position of the second ball in the same time variable as the first. Set the position of each ball expressed in terms of t1 equal to find if they pass.

Thanks! We were making an error in evaluating the time difference in the first antiderivative instead of waiting until the second antiderivative! :) Muchly appreciated!
 
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