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Repeat Problem 26 under the assumption that the rocket is launched from a platform five meters above the ground.

$=40.83 \mathrm{~m} / \mathrm{s}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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here, we're going to be dealing with a rocket that needs to reach a maximum height of 90 meters. We need to determine what the initial velocity ought to be. Have a Maya time. It's going to take to reach those 90 meters. We're further going to suppose that the rocket launches from a structure that is five meters high. Let's start with the differential equation that governs its motion. That would be due to why d t two is equal to G or the second derivative of why is equal to G, which is acceleration due to gravity. Also, since a rocket launches from a structure five meters high, we can say that why it zero or the height at time zero must be five meters on our gold, then is to say, what should the initial velocity be? In other words, what is D Y t. T equal to next will solve the differential equation. By taking anti derivatives, we'll get that d Y. D t on the first step is equal to g times T plus a constant C. We don't have any information on the first derivative for initial conditions, so we'll just proceed to the very next step of taking the anti derivative of what we've just produced. We'll obtain that Y is now equal to G divide by two times T squared plus C times t plus a constant integration deep and that gets us set up. So far, the next thing to do is solve for either one of these constants. By the given initial condition that why it zero is equal to five meters we should be able to solve for D. So let's write this out. So why it? Zero by substitution substitutes in the quantity zero in place of tea so we'll obtain that G over to time. Zero squared plus C time zero plus de is the left hand side, Wyatt zero. But why it? Zero must be five meters, so five is going to be equal to D. And now the equation that governs the motion of the rocket is like walls. G divide by two times T squared plus C times T plus five. For the next step, recall that the rocket will reach its maximum height when velocity is zero. In other words, the max height is obtained when de y duty become zero, but our expression for D. Y. T t is given here, so D Y t T equals zero now means that zero is going to be equal to G Times T plus C. And if we saw for the variable T, then T will be equal to negative C divide by G. So now we know the time at which the maximum height is obtained in terms of the variable. See, we'll be able to use that information to solve for C. But C itself is the initial velocity that we would like to find. So for the next phase of this approach, we go back to this equation that governs the motion of the rocket. And we can right now why, at a time of negative see over G will be equal to through substitution. G divide by two times negative C over G squared plus see times time, which is also negative. See divide by G, then finally plus five, then the trick to this step is negative. See, over G was the time at which max height was reached, but the Max High is also stated to be 90 meters. That means the left hand side of this equation can be written in with just a 90. The next thing we want to do is simplify the right hand side and then solve for the constant C. If we work out, the very first term will obtain C squared in the numerator divide by two G and the denominator plus negative C squared divide by G plus five. So let's do a couple things at this step. Let's subtract five from both sides of the equation, so we'll have a 85 equals down the right hand side of this equation. Let's insert two here as well as here, so that we have a common denominator. Our con women denominators to G. We'll have C squared minus two c squared in the numerator, so just a negative C squared. Overall, if we next multiply both sides of the equation by two G, we'll obtain that 170 times G is negative. C squared. Multiply both sides by a negative one and now negative 170 g is equal to C squared. Now we can solve by taking the square root of both sides. The solution for C is that C is equal to the square root of naked negative 170 times G. But since jeez acceleration due to gravity, we also know that G itself is approximately equal to the quantity negative 9.807 meters per second squared. So negative 9.807 And inserting all this into a calculator tells us that see is approximately equal to 40.83 meters per second. So that tells us next that the initial velocity is de Y t t. At zero. If we insert zero into this expression for velocity, we just get G time zero plus c, and we just found that C is equal to 40 point 83 meters per second. So if we set our initial velocity to that quantity will reach the height of 90 meters. And the problem also wants to know how long it's going to reach that maximum height. To find that require time, we can go back to this expression here and say that the time required is t equals negative. 40.83 Divide by G, which is negative 9.807 and this becomes four point 16 seconds

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