- #1
saturntangerine
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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!
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!