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lara31 [8.8K]
3 years ago
5

A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 50 percent of its previous height. The total vert

ical distance down the ball has traveled when it hits the ground the fifth time is meters.
Mathematics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

0.3125

Step-by-step explanation:

If it decreases by 50% each time it hits the ground then all you have to do is divide 10 by 2 (50%) each time it bounces.

10 / 2 = 5 (first bounce)

5 / 2 = 2.5 (second bounce)

2.5 / 2 = 1.25 (third bounce)

Ect. . .

Until on the fifth bounce it is 0.3125m high.

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Employ a standard trick used in proving the chain rule:

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The limit of a product is the product of limits, i.e. we can write

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The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

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\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
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