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An object that is initially thrown vertically upward with a speed of 2 meters/second from a height of $h$ meters takes 10 seconds to reach the ground. Set up and solve the initial-value problem that governs the motion of the object, and determine $h$

$y=\frac{9}{2} t^{2}+2 t+470.35$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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for this current problem, we have an object that's released with from an initial height of zero meters. We're told also that the initial velocity upon which it'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'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's call this de next. Since we do have initial condition for velocity, let'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'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'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'd also have not yet used the initial condition that the height at time 10 seconds is zero. So it'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'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's word plus two times T plus 470.35 This work then solves this differential equation.

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