Mhm. Okay. What we're gonna do is we're going to walk through a kind of like a word problem. Um We're starting with an object that is projected upward from ground with an initial velocity of 500 ft per second. So this is a vertical motion problem. And the first thing we want to do is we're actually going to be neglecting air resistance. And what we want to do is to determine the velocity equation. And so we know that um the velocity equation generically is our velocity is equal to our initial velocity plus the acceleration times time. Okay, so for our purposes um The initial velocity we're told was 500 ft/s. And of course acceleration because it's vertical motion acceleration is due to gravity, so that is negative 32 ft per second squared. Um So it's going to be negative 32 times T Okay, so there is our velocity function with respect to time and now what we're gonna do is we're gonna actually graph um this velocity function and of course um this velocity function is linear. So this is an equation for a line. But we're gonna go ahead and grab him using a graphing calculator. So I'm gonna switch over to my graphing calculator. Dez knows um is it online, free online graphing calculator. Um And we notice already have it in Dismas. Um And so I'm actually going to um Zoom out. Um and so you notice of course the Y intercept is right up here at 500 and it's a decrease in line. So this is what that decreasing line looks like right here. And now I'm gonna go ahead and quickly sketch my graph. And so if this is T. And this is my velocity Moded at 500. And then somewhere here around 15, a little over 15 is where it crosses. So we're just gonna go ahead and do that. Um Okay, so that is the velocity function. And so now the next thing we want to do is we want to go ahead and use our velocity function to find the position function. And so remember, the position function is equal to the integral of that velocity function. And we're integrating of course, with respect to time. Okay, um and so now we're just going to integrate this. Um and so my position function Is equal to 500 t -30. Open note is not going to be 32, it's gonna be 16 T squared plus some constant of integration. Okay, so now we're going to find that constant of integration. Right? And so what we know is The object is projected upward from ground. So at time zero at a time Zero, my position is on the ground which is at the zero level. And so that is going to be equal to 500 Time 0 -16 times zero Squared. So that is telling me that constant is zero. So really my position function is equal to and I dont know why thats freezing up on me is 500 t minus 16 T squared. And there is my position function. And so now we want to know when at what time do we reach maximum height and what is our maximum height? And so remember, time at max height is when my velocity equals zero. So we're going to set our velocity equal to zero and we're going to sell for that time. So that tells me t is going to be equal to about 15 .625. And so now I want to know what my maximum height is. So I am, my max height is equal to my position function Evaluated at that time of 15.6- five. And of course this is in seconds. And so this will be 500 times 15 0.6 Too far to five. I don't know why that's doing yet -16 Times 15.6-5 squared. And so that will actually be equal. If you put that in your calculator, that is actually going to be equal to 3000 906 point to five ft. So there is my max height. Okay. And now what we want to do is um now we're gonna look and this is if I neglect air resistance. Right? So now what we want to do is um if we're told if air resistance is proportional rips to the square of velocity, okay? Um and it's going to be given by this equation that the derivative of the velocity is equal to negative 32 times. And that should be on the outside. So negative. And in parentheses we have 32 plus K B squared. Okay. Um and we want to use, we want to actually solve this integral equation. So we have one over 32 plus K V squared. D V is equal to negative the integral of D. T. Okay. Um this denominator almost almost looks like an arc tangent. That the problem is this K. That is attached to the U. Squared. Or in our case R V squared. So if I factor out a K. So I'm a factor out. I want over care. I get the integral of one over 32 over K plus B squared. Now that definitely looks definitely looks like an arc tangent now. Um So this integral of DT Okay. Um and so that definitely now looks like one over a squared plus U squared, which is an arc tanne. Right? So um that integral um is a one over Okay. Times and of course the integral um of one over a squared plus U squared is um one over a times arc tangent of you over A. And so that will actually equal negative T. Plus. See okay, so now what we want to do is kind of rewrite this because we want to solve for velocity, we want to have this in terms of velocity. So we're gonna kind of clean this up a little bit um and get velocity by itself. And so we and I do that. Um We end up with this actually becomes one over The Square Root of 32. Okay, and then this will be arc, that's two arc tangent of and then this is the square root of K. Over 32 times fee. Um Equal to negative T. Plus E. Okay. Um and so now we're gonna multiply both sides by that square with just 32. And we're going to solve for T. I'm 40, I'm sorry. And so when we do that we get the is equal to the square root of 32 over K. Times tangent. Because remember we've got to do the inverse of art tangent and C minus. And then we have the square root of 32. Okay. Times T. Okay. Um And now we need to find out what C. Is right. We want to find out what C. Is. And so we want to go ahead. And what we do know is when t when T equals zero, then the Is equal to 500. So if I use that concept and save 500 is equal to. And then We put in here the 32 over K times tangent. So that's just going to be equal to tangent of. See because T. is zero and so we get C. Is actually going to be equal to Yeah. Arc arc tangent. I don't know why this is messing up. Arc tangent of 500 times the square root of K. Over 32. Okay, so now if I put that all back in here we have the velocity as a function of time is equal to Yeah, Is equal to. And then we have this big old long thing, we've got the square root of 32 over K. Times tangent of C. Now is arc. Mhm. Are tangent of 500. Okay. over 32 and then that is going to be minus The Square Root of 32. Okay, times there we have it. That is a big old long crazy mess. Um And then what we want to do is we actually want to um use a graphing calculator. And we're actually gonna graph, we're actually gonna graph mhm. V. F. T. When K. is equal to .001. So we're actually gonna put K. In there as 0.001. And we're actually going to graph VF. T. And from that we actually want to approximate the time zero in which The height this maximum. Okay so wherever there's AK. We're gonna insert .001 and we're going to graph it. So I'm gonna switch back to um my online graphing utility. I'm gonna turn off that linear function. And you notice I already have um this function here right here. And I put in um Where there was, okay, I put in .001. And so I end up getting um I end up getting let's see um 32,000 square to 32,000 times tangent of art can um of 500 times 5000.3125 minus that 0.32 times the X. Value for the T. Value. And here is here is that graph and I'm going to be focused on this section right here. And so you notice that where my height is going to be maximum is the area under the curve right? Um And it's going to have to be a positive area under the curve. And so if I just kind of look at this graph, if I keep blowing up this is kind of like a tangent function. So I know where that area is going to be. Maximum is in this little triangle spot right here from zero to some X. Value. Or in this case t. So that is telling me that maximum height is going to occur when t. Is equal to 6.8 six. So if I switch back over here whips where'd it go? So if I switch back over to my right board. Okay. So um so when K equals .001 um max max height will occur at T equal to 6.86 seconds. Okay. Um and now what we want to do um is used an integration capability to actually find that max height. So we're actually going to integrate from 0 to 6.86 of this function right here. Okay. Um and of course um with K Equal 2.001. And so we're going to actually integrate and we're not gonna do it by hand. Of course that would be totally crazy. So we're gonna do um integrate of I think it was mm 32,000 then it was tangent of are tangent Of and I think it was 500 times the square root of point 1234 and I think it was 3125. And this was all in the arc tangent minus the square root of .03 two T. T. T. Okay, so we're actually going to use the graphing the capability um integral capability to integrate. Um I already have this in here so I'm actually I'm going to I'm just actually changed this and I'm going to do in control from zero. That's the neat thing about gizmos 8 6. And then of course I've got to add in my um the X. Here. Okay and that is saying my max height is 1088, so 1,088 ft. So based off of the integral, so this is about 10 88 feet. And so why is there such a big difference between what we found in part a versus what we did here? And the big difference is in part a or the previous owner where we neglected air resistance. Okay, so in in the previous part, we had no air resistance. And remember, air resistance is going to impede the vertical motion of your object. And so since this latter part, we actually inserted air resistance, we would not expect the object to travel as high because of the resistance of air resistance, the resistance of the air friction.