an-object-is-released-from-rest-at-a-height-of-100-meters-above-the-ground-neglecting-frictional-for

Question

An object is released from rest at a height of 100 meters above the ground. Neglecting frictional forces, the subsequent motion is governed by the initial-value problem
$$
\frac{d^{2} y}{d t^{2}}=g, \quad y(0)=0, \quad \frac{d y}{d t}(0)=0
$$where $y(t)$ denotes the displacement of the object from its initial position at time $t$. Solve this initial-value problem and use your solution to determine the time when the object hits the ground.

Michael Jacobsen · Accepted Answer

in this problem, we have an object that&#x27;s released from rest when it&#x27;s 100 meters above ground. The differential equation that governs its motion is due to white divide by DT equals G orgies acceleration due to gravity. Why, which is the position of the object at times zero is zero and D y t t at time zero, which is the initial velocity is also zero. In order to solve this differential equation first we take an anti derivative of the starting equation. We obtained the derivative of Why So d y d t equals g Times t plus a concert of integration, See, or just in order to solve for this constant of integration? Let&#x27;s use first the fact that the White ET at zero is zero that would tell us that zero equals g time zero plus the constant c. You know the words C is equal to zero itself. Substituting that information back into the differential equation gives us d y t t equals G Times t. Next, let&#x27;s take a further anti derivative. If we take the anti drift of the left hand side, we now have that why is equal to G over to times T squared, plus a new cost of integration. Let&#x27;s call this one D. Now we use the second initial condition that Y zero is equal to zero to solve for that constant de so we&#x27;ll have. Zero is equal to G over two times zero squared plus de. But this implies immediately that D is also equal to zero. So the solution to our differential equation is that why equals g over to t squared for the second portion of this problem? We want to determine when does this object hit the ground to determine this particular time? T recall that the object was dropped at a distance of 100 meters. So we&#x27;re trying to solve at what time tea is the distance travelled equal to 100 meters? Exactly. We have an expression for white t already. So it&#x27;s substitute the right hand side of this equation. We&#x27;ll obtain G over to Times T squared is 100. If we multiply both sides now by two over g, the result is that T squared will be 200. Divide by that acceleration due to gravity G. Upon taking square roots, T is now equal to the square root of 200 divide by G, which is approximately nine point 807 meters per second. And so this quantity turns into approximately 4.5 seconds. So in summary, this expression solves the differential equation, and we know it that in about 4.5 seconds the object would hit the ground.

Michael Jacobsen · Answer

in this situation, we have an object whose initial height is given to be H meters or, in other words, why it zero is equal to H. The initial velocity is given to be V not Nor the words D Y t t f zero is be not. And we&#x27;re let letting t not to know the time when the object will hit the ground. Or, in other words, why a t not is zero. So given that differential equation that governs the motion of this object is second derivative, a position equals G will solve this differential equation. The first anti derivative is de y DT Expression for a lot for velocity is equal to g times T plus a constant of integration. See, let&#x27;s all for this constant immediately were given an initial velocity d Y d T f zero is equal to be not so. In other words, de y d t at a time zero by substitution is G time zero plus C from the equation we just found If we input V not by substitution onto the left hand side, we get to be not is equal to see. So in the words see itself as an initial velocity and the equation is now de y d t equals G Times t plus V not Let&#x27;s take a further anti derivative two obtained that why is equal to G divide by two times T squared plus V not Times t plus a new constant integration. Let&#x27;s call this one deep. Now we go to the initial height for initial condition, which will allow for us to solve for the constant D Immediately. We&#x27;re told that toe y at zero is equal to age. But first, let&#x27;s substitute in zero. In place of the very bull T we obtain G over to time zero squared plus V not time zero plus deep and that zero the height is equal to H. So why it zero is h by substitution. And now we see that H is equal to d. Finally, then we have that Our equation for the motion of this object is why equals g divide by two times t squared plus V not Times t plus H Now Ray To obtain the final portion of this problem, recall that t not was meant to be the time the object hits the ground in other words. Why Tina zero If we express why a t not by substituting in Tina from t in our solution will obtain that g over to times t not squared. Plus the initial velocity V not Times t not plus h on the right hand side but on the left hand side is also zero sore equation becomes zero is equal to do you divide by two times t not squared plus v not times t not plus age. And our objective in this problem is to determine what the initial velocity V not must be in this situation. So it&#x27;s solved for V not in this problem. Let&#x27;s take us our next step. Subtract H and the G over to t not squared from both sides and we&#x27;ll obtain negative h minus g over to t not squared is equal to V, not times t not next. We can say V not is equal to negative H minus 2/2 times t not squared. Almost applied by one over t. Not if on our next step we factor out a negative two from inside the parentheses and reorder the one over Tina factor. We could say that v not is equal to one over negative two times t not times H plus G over two times t not or rather, to h plus g times. T not with the power of to on the variable t not, and this solves for what the initial velocity must be nor to satisfy these initial conditions.

Katelyn Chen · Answer

clear. Ask to show that the following functions trip So I start off. We can find We know the M T is equal to negative 9.8 meters squared meters per second squared. And so then we know that 50 is in a derivative of a hefty which will get us nine nega 9.8 plus a constant. And so then we know that the city is anti derivative Viotti which will get us negative 4.9 t squared plus res ear of tea plus Sarah So that&#x27;s our constant. Now we know that V A T squared is equal to no denying point. Puts read to the do you know squared and so we can actually factor that out to get us negative 9.8 Like the man 0.80 plus reaches zero squared is equal to maybe 9.8 t plus beaches Air It was 99.80 plus of peaches Oh, remember we&#x27;re finding V A t squared from this expression over here Salt scroll down low but to give us a little bit more room. And so now what we can do is we can for you it out to get us naked. 96. Hey sakes. 0.0 for T squared minus 19.6. Reserve T plus three squared of it&#x27;s p zero which will get us so we can rewrite their size. Me 20 My next 19.6 maybe 4.9 t squared plus reaches a roti. And so we know that a city is equal to what we just found earlier than before. 0.9 t skirt plus feet zero t plus us +20 And so if we rewrite it, we get a sooty. My s s zero is equal to negative. 4.9 t squared. He squared plus reaches zero teak in. So based off of that information, we can plug it right back into our original function to get us. So we have about right back into three t squid. We can replace it with this section over here, which gets us be of tea. Squared is equal to you. Squared zero minus 19.6 s o t. My next guest is there

Carson Merrill · Answer

So for these problems, we know our initial velocity, and we know that an object is fired vertically upward and we have its initial position. So we want to find the position and velocity functions for all times. Um, So what we do is we have our velocity function, and then if you take the, um, if we have an acceleration function, that is, that&#x27;s what we&#x27;re going to start off with. So we have our acceleration function with respect to time. When we take the integral of that with respect to t, we end up getting our velocity plus c Well, here&#x27;s the issue. This plus C is causing issues. So in order to get rid of plus C, we need to use our initial condition. The fact that V of zero equals V not means that now we have is, um, RV of T is equal to, um RVF T is equal to this. Right Here are acceleration function plus c. Well, when we plug in zero where we end up, getting is just V, not Equalling. See? So since C is RV not value now, we have v f t equals our function plus v Not then we can take the integral this with respect to t. And then what we end up getting as a result is our position function. So it&#x27;s S f t. And then again we want to use our initial condition because it&#x27;s gonna be plus C so we would have our function and then plus r s not and remember that in this case, Vienna is 29 4. Esna is 30 m, Um and we use all this information to find the highest point. If we look at the position function, we look at the trajectory, we see that right there is the highest point. Um and then we can determine that based on the fact that the slope of the tangent line is zero. So if we set velocity equal to zero software T, we get our value. And then at that time we plug that back into our velocity equation where v f t equals zero and we end up getting that height right here

Linh Vu · Answer

we&#x27;re No in a few. Six time, huh? Freefalling under object doesn&#x27;t depends on time. So it were there would touch the I can&#x27;t understand him on and one way to No can. This question will be in literature craft. The ferocity wasn&#x27;t a 10 year university, and the 10 0 this bit will be maximum will be preserved as mia. And then they will be reducing all of that time. And then after that, desperate will be increasing. Now you know how to mix one. Yeah, I just have never breathe. That team is your team know, wouldn&#x27;t you? So and then the average my Semitic be considered printed on the verge. T jenny constantly is only vomit you.

Averell Hause · Answer

So we&#x27;re going to say downwards is positive And if this is the case, the acceleration would be equal to positive G where we can save positive 9.80 meters per second squared. We can say t minus T prime is equaling 1.0 seconds. Why? Minus y prime is equaling 0.50 h and we know that wise equaling h So we can say why promise? Simply 0.50 h and t itself would be the value of time when it lands. And then, of course, T prime is one second prior to this. So for party, we can say that Why? Prime is equaling 1/2 g t prime squared. Ah, so we can say t prime would be equal to the square root of two. Why prime over g ah, why simply would be equal to 1/2 g t squared and so t would be equal to the square root of two. Why over g now, given this, we can plug in wife H for why, and then 0.5 h for y prime and divide the two equations. So essentially we can say that t t our prime over tea. Ah would be equal to the square root of two times 0.50 H over G divided by, uh to H over G. So essentially, t prime over t is equal in the square root of 0.50 We know that. Ah, we&#x27;re going to say that here t minus 1.0 is gonna be equal to t times 0.50 And this is from knowing that tea prime would be equal to t minus 1.0 So essentially, tee is equaling 1.0 divided by one minus the square root of 10.50 And this is going to yield 3.41 seconds. Of course, T prime would be 2.41 seconds. And, uh, this would be our answer four part. Now we&#x27;re going to plug this results in for why? Why equals 1/2 g t squared. So this would be 1/2 times 9.80 meters per second squared multiplied by 3.41 seconds. Quantity squared. This is equaling approximately 57 meters. This would be your answer for part B and then for part C um, we did not use the quadratic formula because we were able to essentially, um, eliminate this. But we did choose a route because when we said that T prime over T was equaling 0.5 of the square root of 0.50 we said that the square root of 0.50 is positive 0.707 So by saying this, we have actually just chosen the positive route automatically. However again, if we were to choose a negative route, time would then be negative. So here. Ah, if we were rather if we were we time would have to have him in negative point. If we chose negative 0.0.7 would be one minus point Ah, 071 rather one minus negative points. There&#x27;s 707 which would have given us approximately. He would have been approximately 0.6 seconds if the square root of 0.50 was equaling negative 0.707 Now, the thing is is that if this was ah true, t prime would have been negative. So t prime would have been t minus one or 10.6 minus one. So essentially, t prime would have, uh, equated negative number and we can&#x27;t have a negative time. So negative time. It does not exist. So positive route was chosen. That is the end of the solution. Thank you for watching.

Problem

A five-foot-tall boy tosses a tennis ball straigh…

Problem 24 Hard Difficulty

Answer

$y=\frac{9}{2} t^{2}$
$t=\sqrt{\frac{200}{9.807}} \cong 4.5 \mathrm{sec}$

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$y=\frac{9}{2} t^{2}$$t=\sqrt{\frac{200}{9.807}} \cong 4.5 \mathrm{sec}$

Differential Equations and Linear Algebra

Catherine R.

Heather Z.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

Catherine R.

Heather Z.

Samuel H.

Joseph L.

$y=\frac{9}{2} t^{2}$
$t=\sqrt{\frac{200}{9.807}} \cong 4.5 \mathrm{sec}$

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra