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Yo Gendouman, need a little help again.
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A 2.7-kg ball is thrown upward with an initial speed of 20.0 {\rm m/s} from the edge of a 45.0 {\rm m} high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 {\rm m/s}. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored. How far does the woman run before she catches the ball? Thanks man. |
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Re: Yo Gendouman, need a little help again.
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I'm not gendou, but I hope I can help. Since the woman is running at 6 m/s, the horizontal component of the ball's velocity is also 6 m/s. Using Pythagoras we can find the vertical component is ~19.1 m/s. There's that little equation, t = (-v0 +/- sqrt(v02 - 4*1/2a*-d))/a, which is d = v0t + 1/2at2, solved for t (quadratic formula). Just plug in the numbers and we find that the time it takes for the ball to hit the ground is ~5.6 seconds. This is also how long the lady was running, so at 6m/s, 5.6s is 33.6 m. I hope I didn't screw up with anything, I solved for t in my head.
10 more years!
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Re: Yo Gendouman, need a little help again.
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by kudoushinichi
on 2008-02-02 00:19:13
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"How far does the woman run before she catches the ball?" I don't understand this part. If she runs away from the base of the cliff, she'll never catch the ball at all. Unless she goes around in circles... I do have an idea on what the question actually is asking, but I'd like to see how far rob has got before he post this question. And I think fourier understood the question wrongly... or maybe rob wrote his question wrongly. Then again, I could be wrong myself. 8D
Shinjitsu wa itsumo hitotsu!
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Re: Yo Gendouman, need a little help again.
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by
on 2008-02-02 10:54:40 (edited 2008-02-02 10:55:14)
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fourier and kudoushinichi are both correct. ha! 
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