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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$$v_{0}=\frac{1}{2 t_{0}}\left(2 h-g t_{0}^{2}\right)$$

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

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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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'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'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'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'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're told that toe y at zero is equal to age. But first, let'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's solved for V not in this problem. Let's take us our next step. Subtract H and the G over to t not squared from both sides and we'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.

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