4.5.3

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Instructor: So example four, we have-- in this question, we have, a racquetball is hit straight upward with an initial velocity of 2 meters per second. And I'm going to display that, because that is a velocity. Actually, that's an initial velocity, so let's leave that alone now. That's the not equals 2. The mass of the racquetball is this. Air resistance acts on the ball with a force numerically equal to 0.5v, and where v is the velocity, at time, t. Now, find the velocity. 

So fitting this to that form, OK, m times dv/dt, I'm going to say this is 0.0427 times dv/dt equals negative k. All right, well, that's the negative 0.5v minus our mass, 0.0427, times acceleration due to gravity. And I'm going to go ahead and divide by 0.0427 so that we can get this in the right form. You want it to be in standard form. 

So that would be dv/dt equals-- now let's see. Negative 0.5 divided by that would be negative 11.7096v. Now, we actually want it to be in standard form. We need this term to be on the left side. So I'm going to go ahead and adjust this and make that a plus that equals negative 9.8. OK, it's in standard form now, with a p of x equal to 11.7096v and a q of x equal to negative 9.8, which makes our integrating factor-- OK, let's see, mu of x equals e to the antiderivative of 11.7096v. And we're going to go the dt. 

So this would be-- or actually, p does not include that v. OK, that makes a little more sense. So this is e to the 11.7096t. Again, we're disregarding our plus c for the moment. Now, we want to multiply both sides of our differential equation by that mu. So this will be e to the 11.7096t, dv/dt. And let's be consistent with coloring. 7096v e to the 11.7096t times negative 9.8. 

All right, on the left side that is going to fit into the form-- our product rule form, as will always happen with these. t times 11.7096v, and that equals negative 9.8e to the 11.7096t. Now, taking the antiderivative-- oh, first let's write this as the derivative with respect to t of e to the 11.7096t times dv/dt equals negative 9.8e to the 11.7096t. 

Now, we can take the antiderivative of both sides. So that would be e to the 11.7096t times v equals-- and then adjusting for our constant, let's see, that would be-- hold on. Negative 9.8 divided by 11.7096 would be negative 0.837-- 0.8369, we'll do that. 0.836e to the 11.7096t. 

Now, we're going to divide both sides by that e to the 11.7096t to have v equals negative 0.8369-- oh, I left off my integration constant, didn't I-- plus C-- now, that seemed too simple-- e to 11.7096t. 

Now, we're going to use our fact that v of zero equals to the initial velocity to come up with this equation. We have 2 equals negative 0.8369. And evaluating it at zero, we get just C. So plus C, so that C equals 2.88369, so that our particular solution, v equals negative 0.8369 plus 2.8369 over e to the 11.7096 represents our velocity at time, t. 

Now, that's the heavy lifting. Now we do some-- have to do some algebra for this next question. How long does it take for the ball to reach the maximum height? Well, the maximum height occurs when the velocity equals zero. So we'll take that function we just found, 8369 plus 2.8369e to the 11.7096t and said it equals zero. 

And so we can add that negative 0.8369-- we can add 0.8369 to both sides. And I'm actually going to go ahead and write this with a negative exponent, only because it's convenient. Dividing by the 2.8369 you will get-- let's see, that will be e to the negative 11.7096t. Dividing by that we get 0.295, approximately. 

We'll take the natural log of both sides, which will leave us with negative 11.7096t equals natural log of 0.295. I don't want to round any more than I have to. So this would be-- the t is approximately natural log of 0.295 divided by negative 11.7096, which is approximately 0.104 seconds. 

And for our last question, if the ball is hit from an initial height of 1 meter, how high will it reach? Well, we're given a height value, and we're trying to find height. So what we need to do here is actually notice that the derivative of height, or position, is velocity. So we need to actually find the antiderivative. 

So h of t is the antiderivative of v of t, which will be the antiderivative of minus 0.8369 plus 2.8369 over e to the 11.7096t, dt. Oh, and also notice, we have the initial height. So let's-- we'll need this in a moment. h of zero equals 1. 

Now, finding the antiderivative of that isn't too tricky. We don't need integration of our parts or anything like that. We just use our integration rules so that we have negative 0.8369t. And then we will have to divide by negative 11.7096. 2.8369 divided by that value will be minus 0.2423e to the negative-- I'm going to go ahead and leave it like that, again, because it's convenient-- plus C. There's our h of t. 

Now, if we evaluate this at zero, it should equal 1. So we have 1 equals-- let's see, that would be zero-- minus 0.2423 plus C. So the C is 1.2423, making our function h of t equal negative 0.8369t minus 0.2423e to the negative 11.7096t plus 1.2423. 

Now, we need to know the height when the maximum height occurs. Well, we actually know what time that occurs at. So what we need to do now, a last step, is find h of 0.104. I'm going to leave that up to you and a calculator, but what I have found is that it's 1.08357. And that would be in meters.