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Solving for Time and Height of Sphere Meeting | 1 D Motion Problem
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[QUOTE="hjelmgart, post: 4596115, member: 487474"] How do you interpret the "0.5 seconds later". So is the second sphere 60 m from the first sphere, when it is dropped, or is it 60 m from the sphere after the 0.5 seconds?Nonetheless, I do not get the same result as your teacher. I put xf=xi+vit+1/2at^2 For sphere 1 equal to that of sphere 2, since we want to know the time, where xf_1 = xf_2 I calculated a new initial velocity and a new position for sphere 1 after 0.5 sec with: vf=vi+a*t = 25.1 m/s Then the position xf = (vi+vf)*t/2 = 13.775 m xf_1 = xf_2 xi+vit+1/2at^2 = xi+vit+1/2at^2 13.775m+25.1m/s t+1/2(-9.8m/s^2)t^2 = 60m+0t+1/2(-9.8m/s^2)t^2 solving for t gives 1.84s I don't see, how this is wrong, perhaps someone else gets the same? [/QUOTE]
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Solving for Time and Height of Sphere Meeting | 1 D Motion Problem
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