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When the velocity Olan object = large tho magnltudo the force due aintesistancc proportiona v2 with the force acing opposition the molion ol Ihe object_ A shell ol ...

Question

When the velocity Olan object = large tho magnltudo the force due aintesistancc proportiona v2 with the force acing opposition the molion ol Ihe object_ A shell ol mass shot upward Irom Ine ground witn initial velcity of 800 msec the magnitude the force due resistance (0.1)v2 Kanen will the Ghcl reach its maximum height above the ground? What is the maximum height? Assume the acceleration gravily 81 m/s2 .When wlil the shell reach its maximum height above the ground?The shell will reach maximum

When the velocity Olan object = large tho magnltudo the force due aintesistancc proportiona v2 with the force acing opposition the molion ol Ihe object_ A shell ol mass shot upward Irom Ine ground witn initial velcity of 800 msec the magnitude the force due resistance (0.1)v2 Kanen will the Ghcl reach its maximum height above the ground? What is the maximum height? Assume the acceleration gravily 81 m/s2 . When wlil the shell reach its maximum height above the ground? The shell will reach maximum height after 2.65] seconds



Answers

When the velocity v of an object is very large, the
magnitude of the force due to air resistance is proportional
to $$v^{2}$$ with the force acting in opposition to the motion
of the object. A shell of mass 3 kg is shot upward from
the ground with an initial velocity of 500 m/sec. If the
magnitude of the force due to air resistance is $$(0.1) v^{2}$$
,when will the shell reach its maximum height above the
ground? What is the maximum height?

In this problem, we are asked to find the maximum height of a 3 kg shell shot straight up into the air with an initial velocity of 500 m per second. And we're also asked to find the time at which that maximum height occurs. So this is a second law Newtonian mechanics question. So that says the Net force is equal to the mass times the acceleration. So let's start off the problem with, um, a force diagram. So it's shot from the ground. So that's where we would say X zero X being are displacement and then would have going up being in the positive direction. So we have our shell being shot upward and we ask ourselves what kind of forces are acting upon it? We have gravity, which is, uh, pulling the force shell down. So I will have some G and we also have air resistance and air resistance acts contrary to the motion. So since we're only interested in the upward motion since we're only interested in, when does it make it make it to its maximum height and what that height is? That air resistance force is always going to be downward in the situation we're dealing with If we were interested in the entire trajectory. Once the shell starts falling, see Eric resistance will be and upward first force. So the second law says that the net force is equal to the mass times acceleration. So can we compute with these two forces are so the gravitational force is equal to the mass times the acceleration due to gravity which we call g which since we're dealing with, um metric units, it's 9.81 m per second squared and we said that this forces downwards. So we have a negative force and then for our for us due to the air resistance from the problem, we're told that when the velocity is high, the force of the air resistance is proportional to the square of the velocity. So we have a portion Haliti concert Constant V. I'm sorry, okay. And then it's proportional to the square and in the problem they were they told us that the constant is 0.1 and again this force is downward. So we ask ourselves, what is the net force? Well, it's the some of the forces acting on the shell, So we have negative mg minus K V squared and just simplifying a little bit. This is negative quantity mg us K v squared. So this net force is equal to the mass times the acceleration. And we will write acceleration as the derivative of the velocity so that we have a first order differential equation. So we have our net force Negative quantity mg Ask a B squared equal to our mass times acceleration, which is a d v d t. To solve this question, this differential equation. We, um we'll use the fact that it's a separable equation so we can get all the variables to one side and the T variables to the other. We do this by simply dividing by the left side and then getting the DT on the right. So E t is equal to negative m over the quantity mg us. Mm. Where times K t. V Yeah, now that we have our variables separated, we integrate both sides and on the left yet simply t us see our integration constant. Then we're left with a a messier sort of integration on the right side. So the first thing we notice is that there's this plus sign. So if there is this if it was a subtraction sign, we could maybe do, um, partial fraction decomposition to find our and a girl. Since it's a plus sign, we're gonna have to revert to, um, treating the metric integration. So the integral of one over alpha squared plus V squared is one over alpha arc tanne of the over Alpha. So that's going to be the integral that we're going to to focus on. So what we're gonna do is we're going to pull the negative amount of the numerator, and then we're gonna pull this K out of this term by factoring out a K and pulling it out of the integral. So we're then left with negative m over Okay, mg over. Okay, plus B squared. And now we'll insert a change of variable, not a change of variable. A shorthand notation for mg over K quantity squared. And we'll say that this is the square root of mg over. Okay, So what is our and a girl? Now we have negative m over K, the integral of one over alpha squared, plus B squared T V. And this is equal to negative and over K Alpha arc Tangent of the over Al. So now we're interested in finding the So we will try to solve for V So we'll divide by this. And then we'll take the tangent to get rid of the arc tangent and multiply by alpha. So we end up getting the f t is equal to Alpha. I am tangent of negative K alpha T all over. I am plus C. So you might be asking why is there no term with MK and Alpha next to my C? Well, since I haven't computed the C, I can just let the sea absorbed that. So this is a different C than this one right here. But since we hadn't computed it yet, that's a Okay, So this is our velocity function, and we can use our initial condition of 500 m per second. We are initial velocity to say be of zero equals be not so plugging that into here. Yet peanut is equal to alpha times that ancient of C. So solving for C, we get she is equal to the art. Can Vienna over Alpha. We'll continue just writing c to to save, uh, space, But we can now plug that into to get what that is. We can also also find the time of our maximum, Um, hate now. So since we know the velocity and that we've been shot upward and we have a constant force downward, it's going to be traveling up until it has a zero velocity, at which point it reaches its max so we can find our time. I said finding when is V. F t equal to zero t max is when b of t equals zero. So setting the empty quarter zero we have zero equals alpha times with tangent negative k al Fatih over em or seen and we say, When is the function tangent equal to zero and that is actually at zero. So we say, Well, when is negative K alpha t over em see equal to zero and solving for that we get t equals negative C times m over negative K alpha, which is just see em over a alpha. And if we plug in our numbers for this, we get that this is approximately equal to two point six eight seven seconds. So we know the time at which the maximum height occurs, and now we just need to find what that is. And in order to do that, we need to We need to integrate our velocity function with respect to t so called vot is equal to alpha tangent of negative K alpha t all over em, us see. And at this point, we say, Okay, recall that velocity. It's just the derivative displacement with respect to time. So we already have a rather straight forward differential equations. So again, this is separable. So we have the integral of DX is equal to the integral of the alpha times tangent have negative K alpha G over em or C So we have x of t us some constant. We'll call it mhm. And we have to figure out how do we do this integral and well, the Alpha and we recall that sign Ford Co sign is Tangent. And then we need to figure out a change of variables. So if we let Omega equal negative k alpha t over em, plus C and E W is equal to negative. Okay? Uh huh. Over em. DT So with us, have t t is equal to and negative m d w over K alpha. So plugging that in and we noticed that sign and co sign they are related via their derivatives. So pulling all this out, we get negative, M. Whatever. Okay. And a girl now let you echo sign of omega and d you It's negative. Sign of Omega de Omega. So we now can use this negative sign right here. This turns into a positive sign. We have a negative sign over co sign. So we have one over you, Do you? Integrating gives us I am over. Okay. The natural log of the absolute value of you. But you was co sign Omega and Omega was negative. Okay, Alpha t over em plus c. So we can then move that integration constant or to the right. And this is the formula for our displacement. So what we want to do now is find what that maximum displacement displacement is. So we plug in X into X our time where we get the max. And what was that? That time? It was negative. See em over negative K alpha. So plugging that in and we have negative K alpha over em times are C. I'm sorry, Artie, Which was C M over K alpha. There we add C and we had our integration Constant. So we get a negative. See here. Negative C plus C is zero co sign of zero is one so co sign of zeros. One natural log of one is zero. So all we're left with is our integration constant. So the question is how do we find what our integration constant is while we use our initial conditions? So recalled that ex not was the ground right at time zero we were on the ground. So what is our integration? Constant? Well, we have zero is equal to I am over. Okay. Times the natural log, the absolute value of co sign We plug in zero. So all we have inside is C. So what then, is our integration constant? He is equal to negative and over Okay, times the natural log of the absolute value of co sign of C. And if we plug in our values into this we get that he is 101 point 18 seven meters. So our maximum height is 101 0.187 m and it occurs around the time 2.687 seconds

So we know that there's a shell of mass 2 kg so we could have a show right here. And it is 2 kg and it is shot upward with an initial velocity of 200 m per second. And remember, that's initial velocity. That right, there is a good initial condition that we want to keep in mind. We know that the magnitude of the force on the shell due to air resistance is going to be that over 20 velocity magnitude velocity. We want to know when the show reaches its maximum height above the ground and we want to know the maximum height. So, um, what we do is we first want to determine since we have our velocity, The idea is we have velocity, and with velocity, you're able to determine your position. Once we get this position function, the height above the ground is going to look something like this. It's going to be this kind of trajectory. So once you determine that position, you're able to see when the shell reaches its maximum height. It's going to be when the derivative is equal to zero. Now, remember, the derivative of position is actually velocity So in reality, we want to find when the velocity is equal to zero. So once we find that velocity function and set it equal to zero, we can solve for T and we see that that's a T equals 16.48 seconds. And this is through using those initial conditions and everything that we know about air resistance. Then once we have this, we plug this value the time into our position function. So s of 16.48 is then going to give us our height, or Max hype, which is 1536 m for our final answer.

I know friends here it is given that a shared is fire from the ground but the spirit off it emit a per second on at and gets extra degree with Argenteuil. We have to calculate initial or a gentle and vertical relativity I'm taken to reach at the ground. Sorry, I'm taken to reach at maximum this way and see part maximum height. You have to calculate. You have to find the rate that is hell part of it. From the point of projection, it will end. This is just spent. You have familiar this You are meeting Maximize by minus. Why not? An acceleration in our gentle and vertical direction. We start solving from the part it initial velocity in Argenteuil direction. Billy, we not course up 60 in this production We're not is 80 Car 60 is hope so it becomes. Forgive me to Persik by component. Be not sign up 60 Be not a city scientific. E I don't. Three right on solving you will get 63 point. Sorry, 69.30. So these are not the ones airport part of it Hard to be right. You'll read at maximum height. So Ed, maximum height off the projectile. Ever put a gram? It's better city in the vertical direction Become zero. Used a question. We've ice sculpture. We not bite. We lose a valiant duty here. If by is minus Z, this is you know this is given 69.3 minus 9.8 m per second squared so time taken to reach the maximum might you will get 69.3 upon 9.8 on solving it This belly you will get seven point you don't seven. So this is the answer up. Be part thing Maximum height. You can calculate B by square. It's called toe we not by square minus through a by by miners. Why not? So from here by minus by not having the value be very square. We not by square upward, we have I substitute Devon, this is you know on solving it. You will get macron I to be 2 45 m. Okay, Gopal time in here. But is that so hard? Time off flight will be So I soft. I'm taken to reach the maximum. So that is 7107 Order 14.1. Wouldn't sick it so distance covered in Argenteuil direction during the time it is lending on the ground will be hard, gentle velocity and took time taken to reject that route. You may I took time off flight here. Accent acceleration is zero. That's why we can write this. So it will be we not access 40. And this is 14.1 ft on solving it you will get This is Toby 5 55 66 meter. So this is the answer up. Be part and this is the onset off. No. A acceleration in the heart of political direction is zero on in by direction. It is minus. That is 9.8 m per second. This way. Thanks for watching it.

For this problem, on the topic of motion in two dimensions, were told that a shell is fired from level ground at 60 degrees above the horizontal, at an initial velocity of 40 m per second. If it feels no air resistance, we want to find the horizontal and vertical components of its initial velocity the time it takes to reach its highest point, its maximum height above the ground. It's horizontal distance when it lands and its horizontal and vertical components of acceleration velocity at its highest point. So firstly, we want to find its horizontal component of initial velocity V, not X, and we're not. X is equal to we not and the call sign of the firing angle alpha naught. And this is 40 meters per second Times the coal sine of the angle above the horizontal, which is 60°.. And so the horizontal component of initial velocity Is 20 m/s. The vertical component of the initial velocity V, not Y, is equal to the not sign of alpha naught. And this is its initial velocity 40 meters per second. And the white component is multiplied by this sign of 60 degrees. And so this gives us the y component to be 34 0.6 meters the second. So we have the initial velocity broken up into its X and Y components. Next for part B. We want to find the time it takes to reach its highest point. Oh, this object will move in a parabola and at its highest point Its vertical component of velocity will be zero. And so we can use the kind of magic equation V. Y is equal to the note why plus A. Y. Times T. And we can rearrange and solve for the time T. So T. Is equal to the final speed, V. Y minus the initial speed in the Y direction Z. Not why over the acceleration in the Y direction A. Y. Now the acceleration is that only due to gravity. So this is zero minus 34 .6 m/s divided by the acceleration. Which is that due to gravity minus 9.8 m per square second. Since we take downward to be negative, this gives us a time of 3.53 seconds. So that's the time it takes for the shower to reach its highest point. For part C. We want to find the maximum height of the shall about the ground. Now again make it is a kind of magic equation. This time Vienna twice squared or B. Y squared rather is equal to V. Not Y squared plus to a Y delta Y. Which is why minus? Why not? And so to solve for the Y displacement? Why manners? Why not? This is equal to V. Y squared minus V. Not y squared divided by to a Y. That's a speed at the highest point of zero- the initial speed in the Y direction. 34.6 meters per second squared Over two into -9.8 m/km2. So calculating we get the maximum height above the ground off the shell To be 61.2 meters next for part D. We want to calculate the horizontal distance that we shall go, go when it lands. So the total time in the air is twice the time to the maximum height. And so we can use the kind of magic equation X minus X. Note is equal to v. Not X plus a half a X T squared, but there is no acceleration, the X direction since the velocity is constant. So this is equal to the initial horizontal component of velocity 20 meters per second. So actually we know we're not X times T plus a half a XT square. We've forgotten the t Times twice the time to the maximum height, which is two times 3.53 seconds. And so we get the maximum horizontal distance covered by the shell to be 100 And 41 m. And lastly for part E, we want to find the horizontal and vertical components of its acceleration and velocity at the highest point. Now, at maximum height we have V. X. To simply BV narcotics, since the horizontal component of velocity remains constant throughout the journey And this is 20 m/s. At the maximum height V. Y is equal to zero since it momentarily comes to rest at all points in the motion, there's no acceleration in the horizontal direction and the Vertical component of acceleration is constant and that is -9.8 meters per square second. Or that due to gravity.


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