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4s on & !regular training session. When parachutist jumps off an aeroplane his parachute downward with velocity Vo: The velocity of the_ opens; he travels vert...

Question

4s on & !regular training session. When parachutist jumps off an aeroplane his parachute downward with velocity Vo: The velocity of the_ opens; he travels vertically : parachutis and his acceleration is given by at time minutes dv =g-1vwhereacceleration due to gravity and &constant:Show that V =[9M) () Determine the difference in the velocities of the parachutist from the fifth to = the tenth minutes. [2M (c) Find the velocity of the parachutist after a very long _ period. [2M

4s on & !regular training session. When parachutist jumps off an aeroplane his parachute downward with velocity Vo: The velocity of the_ opens; he travels vertically : parachutis and his acceleration is given by at time minutes dv =g-1v where acceleration due to gravity and & constant: Show that V = [9M) () Determine the difference in the velocities of the parachutist from the fifth to = the tenth minutes. [2M (c) Find the velocity of the parachutist after a very long _ period. [2M



Answers

A parachutist is in free fall at a rate of $200 \mathrm{km} / \mathrm{h}$ when he opens his parachute at an altitude of $600 \mathrm{m}$. Following a rapid and constant deceleration, he then descends at a constant rate of $50 \mathrm{km} / \mathrm{h}$ from $586 \mathrm{m}$ to $30 \mathrm{m}$, where he manevers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine ( a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration.

In sector size. We're going to be dealing with the concepts off force and exploration. So before diving into a problem, I'm going to briefly review this concept. So remember that according to Newton's second Law, we have that the sum of all the forces in the body which we called the net force, is equal to the mass of the body times the acceleration. Also remember that the exploration is the first derivative of the velocity with respect to time or the second derivative of the position with respect the time. Okay, so this is all we need in order to solve our problem in our problem. We have a parachutist that has a mess, m okay. And he opens his parachute initially with a speed V 00 And the drag resistance force that the air exerts on the parachute is equal to K. K is a certain constant times V squared where V is the speed off the parachutist at a given moment. Yeah, and our goal given this information, is to find what is the speed via of the parachutist as a function off time t. All right, so remember that the force is equal to M p a. Okay, the squared. Actually the net forces in which we may and the net force on the parachutist is K V squared minus M g. Okay. Ah, actually, I'm sorry. I'm going to use a different convention here. I'm going to consider that this here is the parish parachutist. The positive direction is downwards on notice that the Force the Resistance Force um act upwards while the gravitational force X dollars. So we have f R, which is K V squared. Oh, are better. The weight force, which is a G minus gave you square is equal to M A Now a is DVD t So have M G minus K V squared is equal to m d v d t it. So have the d t is equal to end divided by M G minus K V square. Yeah Devi Yeah, So have the DT is equal to am divided by K times DV divided by m g over K minus V squared. Yeah, and now it can integrate both sides in order to obtain V as a function of time. Now this integral is a little tricky to perform. So let me walk you through it. We have at one over. M do your A K minus V square is equal to one over m g overcame. Actually the square work of mine of M g ever quick k minus V times the square root of them G or K plus b Uh huh. Que en I'm going to try to write it as a divided by the square root of M G over K minus V plus be divided by the square root of M G or K plus V. And you can check that this, uh, equation has a solution. If a equals B equals 1/2 times the square root of K over MGI Yeah, right. Okay, Eso we have that the integral from zero to V of Devi, divided by mg over K minus, the square is equal to the integral from zero to V um D v times one half times square roots of K or M G divided by square root of m g. Okay, Minus um the well, Tess the integral from zero to V, one half square of que over MGI dizzy, divided by the square roots of M G over K plus V right, so we have one half que over MGI times. Yeah, the lot of reason off. Yeah, actually, minus the algorithm off mg over K minus V. Yeah. Yeah. From zero TV plus the logo rhythm of the square root of M G over K plus V from zero to V. Okay, so then another said this is equal to one half off the square root of K over energy time The lover rhythm right off the square root of M G over K plus V divided by square roots of em. Juror K minus Z. Okay, eso now way can go back to our equation. This one here. So we have on the left hand side, we have tea on the right hand side. We have am over k times the integral we just found. So that's one half off the square roots of cave rmg times l n off the square roots of energy over K plus V divided by the square root of M G over K minus V also have t is equal to one half of the square root off M divided by K g l m of the square root of energy over K plus v divided by the square root of energy over K. Mine is he? Oh, so we have to tee times square roots off K you Her m is equal to l n of the square root of them Your cape V divided by the square root of m g K minus v. So I have the exponential yeah of two t times. The square roots of Caju ramp is equal to the square root of key of mgr k mhm plus B divided by n juror key minus the so you can multiply both sides by the square root of them Jew over K minus v So we have the times one plus the exponential off to t times The square root of K G over em is equal to the square roots of NGO ver que times the exponential to g times the square root of Caju Gran minus one. So the speed V is a function of time. Yeah, is the square roots of M g K? Yeah. Times exponential off to tee times square, root of cagey a gram minus one, divided by E. The two t times the square roots of K G over em plus one eso This is the first answer. Then we need to find what is the speed with which the parachute where it should ist It's the ground and that speed must be taken by making t go to infinity. So we have that V of tea. Going to infinity is equal to the square root of m G developed a k. I noticed that this ratio here as t tends to infinity The minus one or plus one will not matter. So this will go to one. So the limit is just the square root of M G, divided by K which concludes their exercise. Yeah, okay.

Problem number 15 Question says that uh question says that But I should drop freely from playing for 10 seconds and before pass it opens out. Okay Then descends with the next eight edition 25 million per second square if the if he builds out the plane at the height 2014 95 m and he has given his velocity reaching the ground will be okay. Now, the last time After 10 seconds, that is before just before opening the parachute will be equal to you. Place GT Initially 0 10 into 10 100 m purse again. Well, if this is a total distance and parachute open here. So let us say as one B. That distance as soon will be equal to As one will be equal to have GT squared because initial velocity V zero so half into 10 and two Dennis Square That is 500 m. Okay, the distance traveled by the parachute under iteration because total it was It is given total distance is being a little hide we should say Is being given at 2495 m. So distance covered under restoration 2495 500 m 1995 m. And if video should be the velocity after reaching the ground will be using video square minus b squared equal to -2 S two because the question is negative. So v dash square will be able to via square minus to wear as to of the 100 sq minus two into Exploration. Reiteration is 25, 1995. Okay, the pain 1000 zero. Okay, Exhibition orientation must have been october 0.5 m per second squared, So this is 900. So in 9000, Which is 25 Covid ish Will be equal to five m/s. The option number is the correct choice. Thank you.

Let Everyone is equals two height of free fall potato free fall before parachute opens. Also at two equals two. Either fall with parachute open ID of all. Then parachute opens. Now we can write, we can right even square. This will be equals two. You square less judy at one. So here We can write it as even square is equal to zero Plus boody at one. Also we can right We can write in zero minus even square. This will be equals two to the attitude. So Let's start with the question one. Any question to this later this week within one Is the question one and this is a question too. Let us add this these two equations. So we will get we will get fuji Edwin is equals two. Who The H. two. Here we need a We need a small correction that that here it will not be G. It will be A. Because their explanation of the parasite is given. It is a. It is a. No we have It's one divided by H. two. This will be equal to any diverted veggie. So this will be called to two divided by 10 or it is supposed to one day wherever five. Now as when we can write total height minus H. Two. So they wanted rights to this will be equals to one divided by are from here we can rewrite this as its two divided by total height adds. This will be equals 25 divided by six. Bye From here we get we get h. two is equal to five divided over six of total ICT capitalist. So this is the correct answer. So now let us take the option. We see that obstinacy is correct.

So the first thing we want to do with this is to determine the velocity for this. Well, if this is our position function T then to get the velocity, we just take the derivative of this with respect to time. So we'll take the derivative of this. So is what unplug essence. And the first thing I'm going to do is pull that 160 out just due to the constant rule. So this would be 1 60 D by D. T of 1 40 minus one plus e to the negative t to fourth. Now we can use the sum and difference rule to distribute this across. So 1 60 and then I also use the constant rules. You want to pull that 1/4? So be 1/4 D by D t t minus d by d t of one plus D by d t of e to the negative t to the fourth. And we could go ahead and take each of these derivatives by themselves. So first, the derivative of tea Well, that's just one the derivative of any constant. To include one is going to be zero and now to take the derivative of E to the negative t four we're going to need to use, um chain. So first it's right out what we have over here. So 14 times one gives us 1/4. Now Chain will remember says we take the role of the outside function, which is going to be e to the T. And so that's just gives us e to the T. And then we plug back in the negative t fourth. And then we take the derivative of what we have on the inside, which is t to the negative forces. And now that there again, we would just pull out the negative one pork, take the derivative of T as within this becomes negative 1/4. So let's go ahead and write all that up. So we have 1 60 oh, 1/4 plus or not, plus minus because it's negative. Negative. 1/4 1 60 he to the negative t fourth. I was just repeat that 1 60 just to make it look a little bit prettier because I don't like to have to look at fractional. This 1 60 is not there. I was getting ahead of myself. Yeah, I don't like to look at fractions if I don't have to So disturbing. That gives 40 minus 40 e to the negative t four. So this is going to be our velocity function based off of Thai. Now, the next thing they want us to do is to show so be they want us to show that our acceleration is going to be 10 minus 1/4 of our velocity. Eso first, let's just go ahead and plug it in over here toe, see what we get just so we can kind of like, check that this is true. So it will be 10 minus 1/4 and then we're taking this and plugging it in for V So 40 minus 40 e to the negative t over four and then we distribute the 1/4. So that gives us 10 minus 10 plus 10, 8 negative teeth. Fourth, which those tens cancel out with each other and we're going to just be left with 10 e to the negative t to the fourth. So this right here is what we want to show, So we haven't actually shown this yet. This is just what we're trying to get to. Alright so to actually get the acceleration the proper way, though, so acceleration is supposed to be the derivative with respect to time of our velocity function Our velocity function we found in the last step. So this will be d by D t of 40 minus 40 e to the negative t to the fourth. And now we can go ahead and distribute this So that gives D by d t of 40 minus So we can use thesis, um, and scaler rule for this over here on the right side, our difference in scaler. So it be 40 d by d t of e to the negative t fourth and again the derivative of 40 0. And then we already took the derivative e to the negative t fourth. But let's just go ahead and repeat those steps again. So there's going to be zero minus 40 so we use changeable. So first we just have e to the negative t forth by itself because derivative eaten X is just eat the X. Then we take the derivative of what was on the inside, which was that negative t fourth and again, This is negative 1/4. And if we multiply everything together. So negative 40 times negative. 1/4 is going to be 10 e to the negative t fourth, so you can see these two are the same. So it checks out now the next thing they want us to do is to find the terminal velocity, which they tell us is just limit as our velocity goes to infinity. So let me look at what our velocity was again. So it's 40 minus that. Okay, so down here we want to do the limit as t approaches infinity of the which again was limit as t approaches Infinity Uh, e or 40 minus already e to the negative t fourth. So if we were to just plug this limit indirectly, this is going to give us a 40 minus 40 e to the negative infinity over, for which that's just going to be negative infinity and then e to the negative infinity. Well, let's go ahead and reciprocate that. So it be 40 minus 40 over e to the infinity and then eats of infinity is infinity. So it be some constant over something going to infinity, which is going to be zero. So this is zero, which means our terminal velocity, which they had losses VR is going to be 40. And actually, do they tell us what units this is in? Um, so it's in meters and our time is in seconds, so actually this is going to be in meters per second. So this is the terminal velocity that we end up with. And then lastly, they want us to determine at what time do we get 95% of the terminal velocity? So that means we're going to set the velocity equal to 90% of this. So down here d time when velocity is equal to 0.9 five VR. So we will do 0.95 times 40 and then we set this equal to 40 minus 40 e to the negative t over four. Now, before I actually simplify this 40 over here, I'm just going to divide everything by 40 because they'll just make it look a little bit nicer. And since we have a 40 everywhere, it was just kind of all cancel out like that. And that's going to give 0.95 is ableto one minus e to the negative t to the fourth so we can go ahead and subtract one over. So it gives us negative 0.5 is equal to negative e to the negative T or four. Then we can multiply each side by negative. So the 0.5 and e to the negative teach fourth. Then we could take the natural log on each side. Uh, any log will do, but natural law will be the most natural one to use. And then that gives natural log of 0.5 is equal tube negative t over four, and then we can multiply over by that negative four. So we get t is equal to negative four natural log of 0.5 Right? And now that we have this time well, actually, let's just see what this is because I mean, I don't even know what that number means. If I were to look at it so natural, log a 0.5 times. Negative four. So this is approximately 11.98 seconds. So we have this here and then the last thing they wanted us to do was to figure out how far we have fallen so Let's go ahead and take this and then plug it in to our very first equation. Um, at the very top here. So let me come back up and pick this up and scoot it all the way down. And then we will need to plug in that value we just got into here. So now to get the distance, like I was saying, we just take this number, plug it in for these teas here. Alright, So s of negative four. Natural log of 0.5 is going to be so it's 1 60 1/4 actually, the 1/4 and that negative or just cancel out. So this, but actually just become negative. Natural log of 0.5 The minus one stays there. And then when we plug that in here, so it would be e raised to the so the negative one in four again counts out. So that's just gonna get us national log of 0.5 and the natural log or eat to the natural log. Cancel that, and that's just 0.5 So this is going to be 160 minus natural log of 0.5 minus one plus 0.5 And then we can combine those to be, um, 1 60 negative natural log of 0.5 minus 0.95 and I'll go ahead and pull the negatives out just to make it look a little bit prettier. So negative 1 60 natural log of 0.5 plus 0.95 And if we were to go ahead and plug this into a calculator because, I mean, honestly, I don't even know what that number is even close to so nuts log a 0.5 loss 0.4 point 95 is going to be about negative to that. We multiply that by negative 1 60 and so this is going to give us something approximately equal to 327.3 meters. So this is supposed to be our how much we fell after we've reached 90 95% of the terminal lawsuit. And if you used a slightly different number like if you would have rounded as opposed to keeping exact um, you might get something slightly different here, but as long as it's within a couple of meters, I would say it's a good chance that you plugged everything incorrectly.


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