Problem

(a) Show that any function of the form $ y = A …

05:36

Problem 54 Medium Difficulty

A model for the velocity of a falling object after time $ t $ is
$ v(t) = \sqrt {\frac {mg}{k}} \tanh (t \sqrt { \frac {gk}{m}}) $
where $ m $ is the mass of the object, $ g = 9.8 m/s^2 $ is the acceleration due to gravity, $ k $ is a constant, $ t $ is measured in seconds, and $ v $ in $ m/s. $
(a) Calculate the terminal velocity of the objects, that is, $ \lim_{t \to ^x} v(t). $
(b) If a person falls from a building, the value of the constant $ k $ depends on his or her position. For a "belly-to-earth" position, $ k = 0.515 kg/s, $ but for a "feet-first" position, $ k = 0.067 kg/s. $ if a $ 60-kg $ person falls in belly-to-earth position, what is the terminal velocity? What about feet-first?

Answer

93.68 $\mathrm{m} / \mathrm{s}$

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Video Transcript

All right, We've got a fun question over here. Excuse me. We got the model for the velocity of a falling object given as to be t is equal to the square root of mass times gravity over K. Excuse me. Yeah, and a k is a constant here. And all of that is being multiplied by the tangent. Age off T square, root of gravity, times constant k over and alright. We know gravity is 9.8 m per second. The rest of these air just variables and we were asked for part a calculate terminal velocity of the object which is just limit t approaches infinity for the function of beauty. Okay, so the way we're going to do this, he is We plug in infinity into our t. We only have one t will notice that deep The value within the 10 age would be so large, Um, that it would essentially go to one. See infinity 10 h t we know is just gonna be one, right? You have a 10 h t. That would just be one. And so you have this Just be equal to M G, okay. Multiplied by one, right? We know our gravity is 98 So we could simple quite a little bit more by saying 9.8. I am over. Okay, Is our terminal velocity good? That'll be part A and Part B wants us to consider a situation where a person is falling from a building and he has a he or she has a belly to earth. Position K is 0.515 but for the feet position he has, she has a point. There were 67 kg per second. And if this is a 60 kg person and it's falling from a belly to earth position, what is our terminal velocity? And then what is our feet? First philosophy. Okay, so we're gonna use are the same equation that we have that we calculated for terminal velocity. But we're just gonna plug in now values. First of all, we know he's 60 kg or she's 60 kg, and then our k we're told for belly first was 55 And if you go ahead and plug that into your handy dandy calculator, you will get a number that is around 33.78 and its velocity. So we know it's gonna be meters per second. Good. And then finally, for part, the second part of B. It's same equation, terminal velocity, same a mass. But this time a different position. And we're told that that k constant points you're 67 So if you plug that into your handy dandy calculator, you will get 93.68 once again at its velocity. So we know it's meters per second. All right, well, I hope that clarifies the question there. And thank you so much for watching.

Mutahar M.

The University of Texas at Arlington

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Problem

(a) Show that any function of the form $ y = A …

Problem 54 Medium Difficulty

Answer

93.68 $\mathrm{m} / \mathrm{s}$

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