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Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of $ 10\;m/s $ and its downward acceleration is$$a = \left\{ \begin{array}{ll} 9 - 0.9t & \mbox{if $ 0\leqslant t \leqslant 10 $}\\ 0 & \mbox{if $ t > 0 $} \end{array} \right.$$If the raindrop is initially $ 500\;m $ above the ground, how long does it take to fall?

$$\approx 11.8 \mathrm{s}$$

03:42

Wen Z.

02:12

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 9

Antiderivatives

Derivatives

Differentiation

Volume

Oregon State University

Baylor University

University of Nottingham

Lectures

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So for this problem we know that rating drops grows, they fall and their surface area is going to increase. Therefore their resistance is going to increase. So we're given this function of acceleration um and we want to determine if the raindrop is initially 500 m above the ground, how long will it take to fall? So we'll take the anti derivative of this since we're given acceleration. So the velocity of this is going to be the anti derivative which will give us a negative 90 plus 0.9 over to he squared. Classy solving for the constant value. See we see that this is going to be minus 10, then um the anti derivative of this is going to be a position function which will end up giving us a negative 4.5 p squared plus 0.15 t cubed minus 10 T plus C, solving for C. Again based on our the fact that we have a 500 m above ground. Initially, that's gonna give us plus 500. So that would be our final function for the position. And then if we look at sF 10, which is the position after 10 seconds um or the in order to follow 100 m, we look at that's equal to 100. Um So it would be where the distance is equal to 400 we see that would take 11 point eight seconds.

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