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

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Two balls are thrown from a cliff 432 feet high, with the first ball thrown at 48 feet per second and the second at 24 feet per second, one second later. To determine if the balls pass each other, it is essential to set their positions equal using separate time variables for each ball. The first ball's velocity function is v(T) = -32T + 48, while the second ball's velocity function is v(T) = -32(T - 1) + 24. The key to solving the problem lies in correctly accounting for the time difference in the antiderivatives.

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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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