an-object-that-is-released-from-a-height-h-meters-above-the-ground-with-a-vertical-velocity-of-v_0-m

Question

An object that is released from a height $h$ meters above the ground with a vertical velocity of $v_{0}$ meters/second hits the ground after $t_{0}$ seconds. Neglecting frictional forces, set up and solve the initial-value problem governing the motion, and use your solution to show that
$$
v_{0}=\frac{1}{2 t_{0}}\left(2 h-g t_{0}^{2}\right)
$$

Michael Jacobsen · Accepted 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.

Michael Jacobsen · 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.

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

Amrita Bhasin · Answer

recall the fact that acceleration a of t is negative. 9.8 in the units are meters per second squared. So we know that velocity is the integral of accelerations of the integral of negative 9.8 d t. Thus, if we integrate, we know the equation would be negative 9.8 times the variable of tea, plus our constant of C. Therefore, we know he can write this as position being the integral of velocity and we can write via, oh, as our constant. You know, its initial velocity. This is negative for 0.9 cheese square plus v a vote times are very bulls t plus c. Then we know, but we can write V F G squared is equivalent to vivo squared. Remember, this is initial velocity squared minus 99 point in terms to 19 point sex. And then this is Asif teed position minus initial position. That&#x27;s us of O

Michael Jacobsen · Answer

for this current problem, we have an object that&#x27;s released with from an initial height of zero meters. We&#x27;re told also that the initial velocity upon which it&#x27;s released is two meters per second and finally that after 10 seconds, the object will reach the ground. Given all this information, our job is to solve for the differential equation that governs this motion. Let&#x27;s take the first anti derivative of this equation well, right. D Y t t, which is the expression for velocity, is equal to G Times T plus a constant of integration. Let&#x27;s call this de next. Since we do have initial condition for velocity, let&#x27;s use that information immediately to solve for the variable we have. D Y t t at zero by substitution is G time zero plus D At the same time, D Y t t at zero is too. So too is going to be equal to D. And now our equation is de y t t is equal to g Times T plus two. Next, if we take a further anti derivative, we get the expression for why is equal to G over two times T squared plus two Times T plus Let&#x27;s use another variable of integration and call it, See if we call the C. Then we could go back to the initial height expression and use that information to solve for C. We can write that the height at time zero or why it zero by substitution from the equation that we just found will be equal to G over to time zero squared, plus two times zero plus c But the same time we&#x27;re also told that Wyatt zero is equal toe age. So we have lower case h is equal to see. And now our equation can be updated to why equals g divide by two times t squared plus two times T plus h, which is our initial height Notice. Next that this differential equation will be fully solved assumes we obtain a value for that quantity. H We&#x27;d also have not yet used the initial condition that the height at time 10 seconds is zero. So it&#x27;s used that information next. So by substitution, weaken first right that Wyatt 10 will be G over two times 10 squared or 100 plus two times 10 plus h. Then at the left hand side is zero equals 50 g plus 20 plus age. And finally, the initial height can be solved by writing. H is equal to negative 50 times. She plus 20. Let&#x27;s find out the approximation for this quality through a calculator. This will now be equal to negative 50 times G, which is acceleration due to gravity and since were in units of meters per second, that quantity G is equal to negative 9.807 plus 20. This tells us that H is approximately equal to 470.35 meters. Next weaken Safer answer that the equation that governs the height of this object is why equals G divide by two t&#x27;s word plus two times T plus 470.35 This work then solves this differential equation.

Katie Mcalpine · Answer

Okay. Um so, Actually okay, we want to get this formula using conservation of mechanical energy on we. Originally this equation 2.11 was obtained using Newtonian physics. So this is kind of neat, e think because it&#x27;s simple air eso you haven&#x27;t object. It&#x27;s not a height. H has initial velocity v ni and vehicles to get the velocity at the bottom. So it&#x27;s of course, going to get bigger because it&#x27;s free falling. So we want to get basically, how much bigger And, um yep, just making sure I got all the given stone properly. Wonderful. So let&#x27;s use energy conservation. There&#x27;s no non conservative forces acting, at least not that we&#x27;re considering. So the initial is equal to the final energy. Initially, there&#x27;s kinetic energy and potential, right? So it&#x27;s initially moving. One happened. Be not squared. Plus, I&#x27;m g each and then finally as 1/2 Um, the final squared. Oh, look at these ums just going away. Goodbye. So then we should do not a different color whatever so that we can get multiplied by both sides by two and take a square root and get BF is equal to be not squared plus two g h. It&#x27;s just so nice toe to see the cohesion between the two constructions

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Verify that $y(t)=A \cos (\omega t-\phi)$ is a so…

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$\frac{1}{-2 to}\left(2 h+g t^{2}o\right)$

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$\frac{1}{-2 to}\left(2 h+g t^{2}o\right)$

Differential Equations and Linear Algebra

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Definite vs Indefinite

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

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Stephen W. Goode, Scott A. Annin

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