an-object-of-mass-5-kg-is-released-from-rest-1000-m-above-the-ground-and-allowed-to-fall-under-the-i

Question

An object of mass 5 kg is released from rest 1000 m above
the ground and allowed to fall under the influence of grav-
ity. Assuming the force due to air resistance is proportional
to the velocity of the object with proportionality constant
$$b=50 \mathrm{N}-\mathrm{sec} / \mathrm{m}$$, determine the equation of motion of the
object. When will the object strike the ground?

Erna Bisschoff · Accepted Answer

So from the given information, we have been given the mass off five. So that&#x27;s let&#x27;s say that&#x27;s M, which is the equal to five and B, which is given as 50. So from this we have the two forces, if one and if to in such a way that if one the first force will be m Times G, which in this case will simplify to five g andi we also have if two which is negative b times the velocity which is a function off t and this will lean simplify to negative 50 times the velocity as a function off t. But we also know that mm times the derivative off the two T is then the force in terms off T and V, and this can become our first equation good. Furthermore, we have that the total force will be if one plus if to which we can substitute with M G plus Negative B VT, which simplifies to five G minus 50 the tea from all the information that was given. And let&#x27;s call this thing equation too. From here, we substitute equation too into Equation one, and we end up with in times the derivative off the velocity to t equal to five g minus 50 V, which is still a function off t. So if we substitute everything indeed it will be five times the derivative off the velocity to time equal to five G minus 50 V. And from here we could divide by five just to simplify but further and we end up with Devi. The derivative off velocity to time is equal to G minus teen the Now we go ahead and we separate the variables. So we put all the V&#x27;s on the one side and the T on the other side, so that will end up to be one over G minus 10 V D. V and we times DT on the other side on the right hand side. And that&#x27;s all we have in terms off T. And from here we integrate both sides. Now we integrate both sides so we have one integral off one over G minus teen v D. V equal to the derivative off the team. So on the left hand side, the integral off the fraction will be negative. 1/10 Lynn off G minus 10 v and on the right hand side that will be t plus the constant C one. I&#x27;m going thio times with the negative tin right through the equation. And I end up with Len off G minus 10 v equal to negative 10 t And if I multiply the C one with negative 10 it just gives me another constant so I can change the c one to a new constant C two From the definition off. Len, this thing can be rewritten as g minus t v equal to e the constant e to the power off. Negative 10 t plus c two. And from this we have G minus 10 V equal to and the E to the power off C two becomes another constant, which I can substitute with C three and we still have to eat to the power off. Negative 10 t Now I&#x27;m going to take the G to the other side and I have 10 negative 10 v equal to C three e to the power of negative T minus g. So this gives me the basic function off the by dividing again with the negative 10. So in other words, the velocity in terms off time will simplify, then to G divided by 10 plus C three e to the power off negative teen T or rather just see because we are. That is our final constant that we are going to work with. Since the object WAAS released from raced, we have the initial condition off V zero with the time is zero and that is also zero. So 40 equal to zero hour V is also equal to zero. And let&#x27;s substitute this thing into the equation that we&#x27;ve obtained here in order to find the value off our constant C. So from that we will have that VT can be substituted by zero. We have G over 10 stole minus or plus C times E to the power off zero that simplifies to zero equal to G over 10 plus c. So therefore, C is equal to negative g over 10. So let&#x27;s put all off this thing back into our VT equation. So we just substitute to see what we got here. So V t is equal to G over 10 Plotnick minus G over teen E to the power of negative tin T and just full simpler. First, we can take out the G over teen as a common factor and we have one minus eater above negative Tain t left just to simplify a bit. Okay, on we will call vesting equation three by integrating the velocity we confined in an equation for X t, which is the distance in terms off the time that was given. So if we integrate this that it will be the integral off VT two DT, which is the integral off G over 10 which is a constant one minus e two part negative team T in terms off t so we can take out the nine the G over teen and we only need to find the integral, then off one minus e to part of negative teen T D. T, which ends up to be G over 10 t plus one over Teen E to the power of negative teen T and remember to add the constant C. So if you look at the situation where it was dropped 1000 m, then we can take our initial distance as being zero and then our distance after time t will then be 1000 so we can take this initial value and substituted into this equation to find the value off the sea. So let&#x27;s do that. That&#x27;s been going to give us zero equal to G. Over. Teen T is equal to zero plus one over Teen E to the power of zero plus C. Let&#x27;s simplify that will be zero equal to G over 100 plus c so they foresee the constant is negative g over 100. Substituting this back into the original equation off X t. We will really integrated. So we&#x27;re going to just simplify this substitute see in the and we end up with X T eggs, which is a function off. T will then be equal to G over 10 times T plus 1/10 e to the power off. Negative 10 t minus G over 100 and just to clean it up a bit, this can be simplified to G over tain t plus G over 100 times e to the power off. Negative 20 minus one. Then finally, if we consider this last bit off the equation, we off the situation where the distance is 1000 m after time T we can use this information to find t so we substitute X t with 1000 thing and we calculate team. So that is 1000 equal to G over tain t plus G over 100 times e to the power of negative 10 T minus one and using the constant G as 981 which is the acceleration due to gravity. We can substitute that as well now, and we&#x27;ll end up with 1000 equal to nine comma 8 1/10 t 10 and then Times T plus 9 +81 over 100 times e to the power off. Negative 10 T minus one. And from this we can simplify to 1000 equal to zero comma 98 won t plus zero comma 09 81 So I&#x27;m just simplifying the fractions into decimals. So that&#x27;s negative. Teen T minus one stole. And from this we get 1000 is equal to zero common 98 won t minus zero comma 09 81 Because when I multiplied the zero comma 0981 with E to the power of negative team, T t is always positive. So it is small enough to neglect. And from here we have seen that 1000 plus zero comma +0981 is equal to zero common +98 won t and that simplifies that T is then approximately one 1019 comma 46 seconds.

Carson Merrill · Answer

So for this problem, we&#x27;re going to set up a diagram, and it&#x27;s helpful to set it up the diagram through the use of a picture. So what we have is, let&#x27;s say, a cliff. And this cliff is 500 ft tall, and we know that there&#x27;s an object. We&#x27;ll just call it this particle object, and it&#x27;s going to be dropped directly from this point down, and we know that this object is £400. So that&#x27;s its force, and it&#x27;s allowed to fall under the influence of gravity. Now, remember, um, the gravity when we&#x27;re dealing with pounds is going to be different. Then when we&#x27;re dealing with Newton&#x27;s. So the gravitational concept constant is going to be 32 0.2 feet per second squared, and based on that, we know we can determine the mass of the object. But more importantly, we want to help determine when the object is going to hit the ground. So the way that we can do that is this right here is an acceleration. So we&#x27;re given acceleration, and our goal is that we want to get from acceleration to velocity, and thankfully, we&#x27;re given, um, some other conditions for velocity such as the air resistance. So we&#x27;ll use air resistance to help us and then from the velocity we need to get to a position function and we want to utilize the fact that we start off with at 500 ft above ground. That&#x27;s going to be an initial condition. So once we get from acceleration, velocity position, we get S of T. This is a function of time is equal to our function, and then there&#x27;s gonna be a plus 500. We find this using the acceleration and velocity. So now what we have to do is plug in. We have to find where this is zero, because when this is zero, that means the height is at zero. So that&#x27;s ultimately going to be what we saw for we&#x27;re going to solve for T. And what we end up finding is that when we set s f T to be zero T is going to equal 13.75 seconds

Carson Merrill · Answer

so it&#x27;s important with these problems to start modeling it, Um, so we&#x27;re gonna have an object right here, and this object is going to have a mass of 8 kg and it&#x27;s going to be given an upward initial velocity. So it&#x27;s going to go up and its initial velocity is 20 m per second and then it&#x27;s going to reach a high point. It&#x27;s going to keep going. It&#x27;s gonna reach a high point, stay right here and then it&#x27;s going to come back down and that is going to be do to gravity. Um and we want to remember that gravity. Since we&#x27;re talking about meters kilograms. Since we&#x27;re talking about those units, we know that gravity is 9.181 m per second squared and in this case, since it&#x27;s going down and if we&#x27;re dealing with direction, we can have a negative acceleration, so then we are going to be able to start with an acceleration. Our goal is to be able to go from acceleration to velocity, and we have a resistance which is going to help us with some initial conditions and making our equation. And then from velocity, we go to our position function with the position function. If we know that the object is initially at 100 m, that&#x27;s going to be an initial condition. We have. So what we have is 100 m, something like this. Perhaps we&#x27;re on a cliff and we throw it up. And then there&#x27;s going to be this 100 m drop in addition to the height that was already gone up. So it&#x27;s going to go all the way down. So we have to keep that in mind with our initial conditions. Then we&#x27;ll get a position function of the form s f t equals, and that&#x27;s what we have to find. So then what we do is we set this equal to zero. We set S f t equals zero. And when we do that, we end up finding our final answer. Um, and what that&#x27;s going to be is that the object is going to strike the ground when t equals 22.9 seconds

Keshav Singh · Answer

this problem on the topic of Newton&#x27;s laws of motion, we have an object that is falling under the influence of gravity and also acted upon by frictional force of air resistance. If we know the magnitude of this force is approximately proportional to the speed of the object so that the frictional force committed as B times V, where we are given the constant B and the mass of the object, we want to find the terminal speed that the object reaches while falling. And then we want to know if this answer depends on the initial speed of the object. So here we have a force diagram showing the frictional force acting upward in the object, which is minus B V and its weight acting vertically downward, which is M Times G, where Emma&#x27;s its mess. Now the object will fall so that, according to Newton&#x27;s second Law, M. A is equal to the weight mg minus the frictional Resistance BV, meaning that if we rearrange the equation, we get the acceleration. A to B mg minus b V over M. M is the mass of this object. If we take the downward direction to be positive, we know equilibrium. We reached when a is equal to zero. So when a is equal to zero, we get the velocity to be terminal velocity and this is simply MG divided by B and this is the mass of the object. 50 kg times the acceleration due to gravity 9.8 meters per square second divided by B, which is 15 kgs per second, which gives us the terminal velocity of this object to be 33 meters per second. So this is the speed which the object cannot exceed. Now we want to know if the speed depends on the initial velocity. Now, if the initial velocity we not is less than 33 m the second, then we have a positive acceleration a greater than zero. But we still have a final velocity the of 33 meters per second. That&#x27;s the largest velocity that the object can attain. On the other hand, if the initial velocity we not is greater than 33 m a second, then we have a negative exhilaration or a deceleration. And 33 m per second is the smallest velocity attained by the object. But it&#x27;s still the final velocity or the terminal velocity attained by the object. So even if the initial velocity is 33 m per second, acceleration is zero and the object continues to fall with the constant speed of 33 m per second. So the answer is no. Our answer in part a does not depend on the initial velocity.

Ankit Gupta · Answer

first we have to find I&#x27;m taken by the object to reach 15 meter off the ground, which means that we take this tense as is equal to 15 meters. We have given that a question for distance, as is equal to minus 4.90 square. Plus I think he so put the value of as here we get minus 4.9 e squared plus 20 e is equal to 15 which implies that minus 4.9 a squared plus 20 T minus 15 is equal to zero. Now take common minus 4.9. From here we get e squared minus 4.8 e plus 3.6 is equal to zero which implies that when we make factory this equation we get t minus one B minus 3.3 is equal to zero. From here there are two values off the, which are a is equal to one and 3.3 So we take is equal to 3.3 seconds. So the time taken by the object to reach 15 meter off the ground is 3.3 seconds. Now we find the time, then object strikes the ground. For this we take distance s is equal to zero. So the equation off distance minus 4.90 square plus 20 p is equal to zero which implies that Van V A minus 4.9 comin from where we get e squared minus 4.0 80 is equal to zero a day Comin from here we get a minus 4.8 is equal to zero. From here we get two values off TV charge zero and four point Zito eight we neglect physical zero we take is equal to 4.8 second. So the time taken by the object to strike the ground is 4.8 seconds. Now we have to find the time when the object Rita height off 100 meter. For this, as is equal to 100 meter saw the equation minus 4.9 T square plus 20 e is equal to 100 which implies that minus 4.9 a square last 20 e minus 100 is equal to zero. Now take 4.9 comin from here we get minus 80 square plus 4.0 it a miner&#x27;s 20.4 is equal to zero, which implies that minus e square minus 7.0 40 minus 3.4 T minus 20.5 is equal to zero now making factors minus e e plus 7.4 minus 3.4 e plus 7.4 is equal to zero we get to l use off a. The chart A is equal to three fine 04 and A is equal to 7.4 we take is thick or 27.4 seconds, which means that the time taken by the object to read the height off 100 meters is 7.4 seconds.

Gideon Idumah · Answer

here were given that ham the courts on jerky legroom. Well, given the G E quarter to 9.8 meets up a second squares, we give up, be equal to five kilogram per second, and they want to find a limiting and terminal velocity. Now, plug in this palace into the equation you&#x27;re gonna have am which is 100 dividend sing. He called to M, which is 100. What supplied by G, which is nine point ends on the moor, replied Sorry. Under minus B, which is five, then times V. Now all we do is divide with side by five. You&#x27;re gonna have that 20 David it see, he called. So this is 1 96 1 soon miners being which I would re writes for simplicity as minus into bracket being minus 1 96.2 Not divide with sights off this equation by the minus one and a six point. So you&#x27;re gonna have a DVD over the minus one and a six point sue he called toe, then establish minus the miners, then divide beside by chance. You&#x27;re gonna have one over 20. What supplied by did see? Now let&#x27;s take the integral of both sides. This will be natural. Low off the upset of L O V. Minus 19612 equal to minus when into goodness I have seen over Trance and then lost one. Now what we do next is with big explanation. Both sides. You&#x27;re gonna have B minus one of the 6.2 Bacall toe exponential minus C. Constraints multiplied by explanation of See, Juan Duck implies that being minus one of six point Sue equal to seek to eating minus 20 seats. Who is a ghost? Oh, need to this one. Now take off the absolute value. RV minus 106. 12 equal toe. See eating Minus Ovitz Wincing. Yes. Scene is water blossom miners soon. That implies that be equal to 106 point soon. Plus seeing eating minus over anything. Now, given that the off zero the quart. Two centimeter per seconds. So I plugged That mean I have off zero, which is 10 equal toe 1 96 horns soon below sea to zero, which is one. This implies that scene also minus one in six point Soon that implies that box of being start like veal scene we call toe one night. It&#x27;s six point soon, minus one. It is 61 Soon eat. Remind us over 20. Now I want to find the terminal velocity to find a terminal velocity you leads. See, go to infinity So that the infinite which is the time, not the loss. It&#x27;s dust limits as the ghost infinity off one at a 6.2 minus one in this 61 to eat the finest of its currency. When see Ghost infinity, this goes to zero. Why? Because you are eating minors. So this comes down. So this goes to zero. So that implies that vegan infinity, which in some velocity is just 1 96.2 seconds.

Problem

A $$400-1 \mathrm{b}$$ object is released from re…

Problem 1 Easy Difficulty

Answer

$t \approx 1019.46 \mathrm{sec}$

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$t \approx 1019.46 \mathrm{sec}$

Fundamentals of Differential Equations

Catherine R.

Kayleah T.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

R. Kent NagleFundamentals of Differential Equations

Catherine R.

Kayleah T.

Kristen K.

Michael J.

R. Kent Nagle

Fundamentals of Differential Equations