Problem

For the following problems, set up and solve the …

Problem 285 Easy Difficulty

For the following problems, set up and solve the differential equations.
You drop the same ball of mass 5 kilograms out of the same airplane window at the same height, except
this time you assume a drag force proportional to the ball's velocity, using a proportionality constant of 3 and the ball reaches terminal velocity. Solve for the distance fallen as a function of time. How long does it take the ball to reach the ground?

Answer

$y(t)=\frac{-245}{9} e^{-3 t / 5}-\frac{49}{3} t+\frac{45245}{9}$
307.79 seconds

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Calculus 2 / BC

Calculus

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

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Video Transcript

So for this problem were given a falling distance equation 4.9 t squared plus V not t equals s So 4.9 times the time value squared, plus the initial velocity value times a time is equal to the distance for the altitude value and there are three parts. This problem in the first part, we're told part A that the distance is 505 100 meter altitude and that the initial velocity is zero since it is simply dropped from a plane where it falls off. So how long will it take to reach the ground? We need to calculate the time here. So time is the unknown value 4.9 times time squared is equal to 500. And since the velocity initial zero, we don't have to worry about this term. Over here, the non times T just adds up to zero. So we're gonna take the T squared, divide out the 4.9 and it equals to 500 over 4.9. Andi, simply take the square root of this value and when we calculate this will get that time is equal to about 10.1 seconds for part b were given, but the distance is still equal to 500 meters this altitude and this time we have an initial velocity off 30 meters per second. So this means that we need to use both parts of the equation now 4.9 times the unknown time value squared plus the 30 meters per second times unknown time by will. Give us that 500 meters altitude. Now we're simply going to solve this with the quadratic formula by bringing over the 500 subtracted. So now we'll have 4.9 t squared plus 30 t minus 500 we consult for two using the quadratic formula. So negative B which is negative. 30 plus or minus Radical B squared. So 30 squared minus four times a ton. See so four times 4.9 times 500. And that's all over two times a so two times 4.9. When we calculate this mostly that we only get one positive answer. And since we need our time body to be positive, we can put this as our final answer of 7.49 seconds. And finally, for part C, we're given a time value of five seconds, an initial velocity of 30 meters per second and we have to calculate the distance traveled on our own. So we have 4.9 times the five seconds squared and then 30 meters per second times the five seconds and simply do the math on this one with a calculator and move. Get 272.5 meters traveled. So just to recap, our final answer for a is 10.1 seconds. Our final answer for B is 7.49 seconds and our final answer first see is two points at 272.5 meters.

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