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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.

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

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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in this problem, we have an object that's released from rest when it'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'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'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'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'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'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's substitute the right hand side of this equation. We'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.

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