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A five-foot-tall boy tosses a tennis ball straight up from the level of the top of his head. Neglecting frictional forces, the subsequent motion is governed by the differential equation$$\frac{d^{2} y}{d t^{2}}=g$$If the object hits the ground 8 seconds after the boy releases it, find(a) the time when the tennis ball reaches its maximum height.(b) the maximum height of the tennis ball.

a. $g \cong-3.98 \mathrm{} \mathrm{s}^{}$b. $259.866 \mathrm{ft}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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for this situation. We're told that a boy that's FIFA tall tosses a ball and releases right at the height of five feet, giving us the initial condition that why it zero is five feet. We're also told that a seconds later the ball hits the ground, giving us this second initial condition. So first, we're going to solve this differential equation. By taking the anti derivative of both sides, we will attain that D Y T t is equal to g Times T plus a constant integration. See, ordinarily, we try to solve for C immediately. But note that both of our initial conditions does not involve a derivative. That means the very next step is to take immediately the next anti derivative, which will be why equals from the anti driven of the left hand side G over to Times T's word plus C Times T plus a new constant integration. Let's call this one upper case D. Now we're ready to go to the initial conditions and Saul for both constants C and D. Starting out with D, we can use the initial condition that Wyatt time zero is five feet. Let's right out Y at zero by substitution allows us to write G over to time zero squared plus C time zero plus de Now we substitute in the value for Wyatt zero, which is five feet and obtained that five is equal to de itself. Bringing this back in our solution is now why equals g divide by two times t squared plus c Times T plus five. Next, we repeat the process with the second initial condition and right that why at eight equals by substitution g over to times eight squared plus C times eight plus five. Now we make another substitution, or why eight become zero itself from the stated initial condition. So now we have zero is equal to the eight times eight is 64. Divide by two gives us 32 times acceleration due to gravity G plus eight times see plus five and recall. Our goal is to solve for that constant C So subtract five and 32 g so minus five minus 32 g from both sides of the equation is now equal to eight c and divide by eight. So negative five. My is 32 g divide by eight is equal to see itself. One thing we should be very careful about this problem is the units are given in feet. That means G can approximate as negative 32.17 feet per second squared. If we make that approximation will find through a calculator that C is approximately equal to the quantity 128 0.55 So it's expressed that next in our solution, we now have that why is equal to G Divide by two T squared minus 128.55 t plus five where that negative came from here and here, which means it should have gone here a swell. So now we have an expression for the motion of this object. We can now answer the question off. When does the ball reach its maximum height? Well, the maximum height will always occur when the derivative is zero. So let's take this equation here and set that equal to zero and solve. So for the next phase we have zero is equal to g Times T plus see itself. But we just found out that C is equal to negative 128.55 so our equation now turns into zero is equal to G Times T minus 128.55 If we saw for time T, we'll get 128.55 Divide by acceleration due to gravity. G is our time. T. Once we put this in, the calculator will be ready to some rights. Are results will say next that the max height is reached when T is equal to approximately. After throwing this expression to the calculator about 3.98 seconds, then the problem asks for a second question. We want to know what the actual max height would have been. So we take the quantity 3.98 seconds, which tells us when the ball reaches its max height and go back to the formula we saw for why, which tells us the maximum height at any time. T Let's evaluate why, at three points 98 which will give us G over to times 3.98 squared minus 128.55 times 3.98 plus five. Upon putting all that work into the calculator, we have that the height at 3.98 seconds, is going to be approximately equal to 259 point 866 feet and this is the max height that was reached.

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