**One stone is falling down, the other is going up, when do they meet?**
The answer is actually asking for the height at which they meet but I know that if i have the interval I can plug it in and get the height but when I attempt to solve it, I get t = 0 which is wrong and would indicate that the first stone lands at the same time the person throws the other stone which doesn't sound realistic in terms of school physics problems. __Here is the question:__
"Suppose you throw a stone straight up with an initial velocity of 27.5 m/s and, 4.5 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?" __My work:__ **Stone 1:**
a = -9.8 m/s^2
Vi = 27.5 m/s
t = 4.5s
Vf = Vi + at
Vf = -16.6m/s
Δy = (Vi)(t) + (1/2)(a)(t)^2 *<---- Here was my mistake! Correct way: yf = (Vi)(t) + (1/2)(a)(t)^2 + yi where I find yi to be 24.525* **Stone 2:**
Vi = 27.5m/s
a = -9.8 m/s^2
Δy = (Vi)(t) + (1/2)(a)(t)^2 **When they meet:**
Δy = (Vi)(t) + (1/2)(a)(t)^2 = (Vi)(t) + (1/2)(a)(t)^2
(-16.6m/s)(t) = (27.5m/s)(t)
0 = 291.6t
t = 0
I would REALLY appreciate it if someone could tell me what I am doing wrong!
Any help would be greatly appreciated!
Edit: I solved it.
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Last edited by s3a; Feb 17th 2010 at 07:00 PM.
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