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anonymous
 5 years ago
I'm working on the 18.01 Part A Problem set Question 1B2 d and e. I'm not sure how to do it... can anyone help?
anonymous
 5 years ago
I'm working on the 18.01 Part A Problem set Question 1B2 d and e. I'm not sure how to do it... can anyone help?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assuming you've already solved a, b and c, you can look at d like this: Our graph of v is going to start at time = 0, let's assume that the x axis is our time, so our first point is going to be on x=0, so now let's find y. y is going to be the speed at which the ball is moving (in this case it's moving up, so it's positive) and we know that at time = 0 it's going the fastest it's going to go in that direction because it's going up and it slows down, we learned that in problem b. We are also told that its initial speed is b. So our graph of v is going to start at (0, b), we aren't given b. We also learned in problem b that v = 0 at b/32. We can also see that the change in velocity is linear so the line would be straight from b to b/32. Another, quicker way to do this would simply to take v=b32t and put it into slope/intercept form. We know that time is our x axis so if you don't immediately see it just replace t with x and v with y and you'll see y=b32x or y=32x+b, which means our graph would start at b and have a slope of 32, and to get the exact x intercept, since we don't know b, we just make y=0 and solve for x. \[0=32x + b\]\[ 32x = b\]\[x=b/32\] The second graph, s, is a little harder. It's got \[t ^{2}\] in it so we can assume it's a parabola. Also based on our previous graph we can see that it will start off with a slope of b and end with a slope of 0, then have a negative slope. We also know that it reaches its highest point at t=b/32 because we've done that problem. We can guess that it will reach 0 at \[2*b/32\] based on our previous graph as well. So our graph will be a parabola starting at t=0, and reach its max height of \[s=b^{2}/64\] at t=b/32, and reach 0 again at t=b/16. As for problem e. Our bounce is half the max height of our second bounce is half of our first. We know the max height of our second bounce is \[b^{2}/64\] and we know that half of that is \[1/2 * b^{2}/64\] or \[b^{2}/2(64)\] I'm not multiplying that 2 into that 64 because that will make it easier on me later. so, let's label our bounces and solve for the initial velocity of bounce 2. \[x = bounce 1, w = bounce2\] \[w^{2}/64 = x^{2}/2(64)\] \[w^{2} = x^{2}/2\] \[\sqrt{w^{2}} = \sqrt{x^{2}}/\sqrt{2}\] \[w=x/\sqrt{2}\] So the initial speed of our second bounce is w = x/ sqrt(2) the only thing that's going to change in this graph is b, and since the value of t relies on b being variable, the graph should look the same, but with a different max height. Since this is the second bounce, we should add the times of bounce1 to our graph, so problem e will start at the end of the second graph in problem d, so start at x/16 and end it at w/16, the highest point being (w^2/64, w/32). I hope this makes sense. Let me know if I made any errors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also, you can treat that sequence of bounces as a geometric sequence or the maximum heights of each bounce as points on an exponential decay curve.
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